CLEP College Algebra/Sequences and Series

In mathematics, it is important to find patterns. That is what mathematicians do almost everyday of their lives. How they determine patterns is different depending on the type of mathematics they work with. For college algebra, determining patterns is part of the curriculum. The problem below demonstrates one way we determine these patterns. Keep in mind that the problem below is an Exploration and most likely not representative of the types of problems you may see in the CLEP College Algebra exam.

What were we doing in the problem above? Essentially, we were simply trying to find what position some "term" is in. Does it not intrigue you to see math try to find a position of some "term" in a list? We have problems like these as mathematicians because the patterns underlying a "list" of numbers can help us determine new facts of mathematics. After all, what were we doing when using functions? We were trying to find a number using a pattern (the function). Unlike the previous sections, however, we were not given a formula. Luckily enough, it is not difficult to make a formula for a given "list" of numbers. Before we dive into these new problems. It helps to establish definitions.

There is a little disclaimer to get out of the way before we try to solve some problems. First, a sequence can have no pattern. However, for our purposes, we will not count any list of numbers in which no pattern exists. Second, even in simple sequences, numbers of any kind can be named if they follow after a sequence rule. For example, here's a sequence in which the rule is to list prime numbers: $\{2,3,5,7,11,\dots \}$ If you were like most people, you would probably name the prime numbers in order. However, you could perhaps finish the sequence like this: $\{2,3,5,7,11,13,23,2011,\mathit{(}{2}^{82,589,933}\mathit{-}1\mathit{)},\dots \}$ For our purpose of standardization, we will follow a pattern by stating what number you must find first or identify the pattern.

Let's begin exploring the world that is Sequences and Series.

Sequences

As you already know, a sequence is a list of objects that generally follow a pattern. However, the type of pattern that is described will classify sequences into either arithmetic sequences or geometric sequences. Each will be explored in depth within the next sections.

Arithmetic Sequences

An example could perhaps help you figure with the formal definition above: The sequence $\{2,4,6,8,10,\dots \}$ has a one-to-one correspondence with the general sequence $\{x_1,x_2,x_3,x_4,x_5,\dots \}$ since the first term $x_1=2$, second term $x_2=4$, third term $x_3=6$, and so on. The difference is the amount added to each previous term to get the new term. For $x_2$, $x_1+d=x_2$ or $2+d=4$. Solving for $d$ is the difference of the two terms. In this example, the difference is $d=2$. This is how we define an arithmetic sequence.

Recursive Formula

Often times, we want to generate a sequence using a formula (we are mathematicians, after all, and we like to study sequences to see if there are any general patterns). If we want to find $x_i$, we may use the following formula:

$$x_i=x_{i-1}+d.$$

However, the above formula could describe any sequence that has that general pattern. To fix this, we need to describe the first term as well when using the formula above. There are two ways to describe this formula:

1. We have it horizontally deliniated: $x_1=a;x_i=x_{i-1}+d$.
2. We have it vertically deliniated: $\{\begin{array}{c}{x}_1=a\\ x_i=x_{i-1}+d\end{array}$

To save space, we will horizontally deliniate formulas for arithmetic sequences in this WikiBook.

Formulas in which the first term is identified along with an equation in which the previous term is added by $d$ to get the next term is called a recursive formula.

Template:ExampleRobox To determine the next term, we first need to find the difference $d$. Note that an arithmetic sequence will have the next term add $d$ to the previous term. Since that is how the arithmetic sequence works, $19+d=18.25$ is a valid way to find the difference between two terms. Solving for $d$, we find the constant difference is $d=-\frac{3}{4}=-0.75$. Since the difference is the same for each given term in the sequence, we can find the 7^th term by adding $d=-0.75$ to $16$, which gives us $16-0.75=15.25$. Finally, add $d$ to the next term to get the final answer:

$$15.25-0.75=14.5.$$Template:Robox/Close

There are many reasons why it is more important to have a recursive formula. It is not always slow; it may be easier to understand. The next example shows why this is exactly true

Template:ExampleRobox Many of you will perhaps know this famous pattern as the Fibonacci Sequence. For those of you who do not know this sequence, the way we determine the next term is by using the previous terms and adding them together. In our notation, we would say that the term $a$ at index $i$, $a_i$ is equivalent to $a_{i-2}+a_{i-1}$. Remember, however, we are not done. If a mathematician saw the sequence $\{2,4,6,10,16,26,42,\dots \}$, he (or she) would determine that $a_i=a_{i-2}+a_{i-1}$ also describes that sequence. Therefore, we must list the first two terms because listing only the first term would not allow us to get the next term. This means our final answer is

$$a_1=1,a_2=1;a_i=a_{i-2}+a_{i-1}.$$Template:Robox/Close

Note that Example 1.1.b is not an example of an arithmetic sequence. Your next exploration will be to determine why this is true. Along with that, you will use your critical thinking skills to argue for or against something in the explorations after that one.

Direct Formula

By now, you may be wondering if there is a way to find the term of an arithmetic sequence directly. Well, there is. Before giving you the formula, let us go through the motions for our general recursive arithmetic formula $x_1=a;x_k=x_{k-1}+d$. Let's chart the recursive arithmetic formula.

$$\begin{array}{cc}k& x\\ 1& a\\ 2& a+d\\ 3& a+2d\\ 4& a+3d\\ ⋮& ⋮\end{array}$$

If you think about it, the table above is basically a linear function, although starting at $k=1$ instead of $k=0$. Write out the function as $x=a+kd$. We get near our answer. Our independent variable $k$ is horizontally translated by $1$ to the right, so $x=a+(k-1)d$ is our function. In fact, we found our direct relationship. Rewrite it the way we normally write it and we found our direct formula: $x_k=a+(k-1)d$.

Template:ExampleRobox As always, before we can determine the answer, we need to find the "rate of change" of our sequence. Since $a=22.4$, we know that $x_i=22.4+(i-1)d$. The second term is $29.2$, so when $i=2$, $29.2=22.4+(2-1)d$. Of course, now we can solve for $d$:

$29.2=22.4+(2-1)d$;

$29.2=d$;

$6.8=d$.

Since we now know the common difference, we can find the smallest $450$^th term in the sequence. By using our direct formula for an arithmetic sequence, we can find the index in which it is possible. Since $x_i=22.4-6.8(i-1)$, we find the $450$^th term by substituting $i=450$. Ergo, $x_{450}=22.4-6.8(450-1)$. Solve for $x_{450}$ to get the final answer.

$x_{450}=22.4-6.8(450-1)$;

$x_{450}=22.4-6.8(449)$;

$x_{450}=22.4-3121.2$;

$$x_{450}=-3030.8.$$Template:Robox/Close

The example above would be a routine, straightforward problem in the CLEP College Algebra exam. However, as practice makes perfect, we will also have non-routine problems that involve thorough understanding of the topic and concepts and skills learned, which will make up 50% of the exam. This is why it is important to do the explorations. While they may not be on a CLEP exam, they are vital in making you think like a mathematician. The next problem will be non-routine problem.

Template:ExampleRobox As always, before we can determine the answer, we need to find the "rate of change" of our sequence. Since $a=22.4$, we know that $x_i=22.4+(i-1)d$. We know that the second term is $21.2$, so when $i=2$, $21.2=22.4+(2-1)d$. Of course, now we can solve for $d$:

$21.2=22.4+(2-1)d$;

$21.2=22.4+d$;

$-1.2=d$.

Since we now know the common difference, we can find the smallest term needed to reach a negative number in our sequence. By using our direct formula for an arithmetic sequence, we can find the index by solving for $i$. Since $x_i=22.4-1.2(i-1)$, we find the minimal index in which it is possible to have a number less than zero. Ergo, $0\ge 22.4-1.2(i-1)$. Now all we have to do is solve for $i$.

$22.4-1.2(i-1)\le 0$;

$-1.2(i-1)\le -22.4$;

$i-1\ge \frac{-22.4}{-1.2}$;

$i\ge \frac{22.4}{1.2}+1$;

$i\ge 19.\bar{6}$.

Since index $i$ must be greater than $19.\bar{6}$, the minimal index required to find the term that is negative is at

$$i=20.$$Template:Robox/Close

CLEP Practice Problems: Check Your Understanding

<quiz display=simple> {In the arithmetic series $\{2,6,10,14,\dots ,398\}$ in which $398$ is the last term, how many terms are in the sequence? |type="{}"} { 100 _9 } terms. (Note: typing a comma will give you a decimal, so do not type a comma.) || To find the number of terms, first find the constant difference $d$ from subtracting the first two values (this is just easier to do). We learn $d=4$. Next, since a direct arithmetic formula helps us find any value of index $i$, we can find the number of terms by using the direct arithmetic formula: $x_i=a+d(i-1)$. Since we know that at some value $i$, we can find term $x_i=398$, we simply use those values into our direct arithmetic formula. Therefore, $398=2+4(i-1)$ can help us find the number of terms $i$, since the last term is $398$. With this, after doing all the math, you get $i=100$.

{A school district has set-up a system whereby each school week day (days Monday through Friday) is categorized into either an A-day or B-day. Abigail is tired of having her pencils not returned to her. She decided to devise a system in which each student will record their first and last name. The pencils are free for anyone to borrow. However, should a student forget to return a pencil to her before the end of class, a student is charged $25$ cents first. After, another $25$ cents are charged per cycled block day the pencil is not returned – e.g. if a student accidentally forgot to give back Abigail's pencil, they will be fined the next A-day after the B-day. Joseph had forgotten to return the pencil for $60$ school week days. How much does Joseph owe Abigail, according to her system? |type="()"} - $$15.00$ || While a student may have used the correct formula, the student did not account for block days. Since there are two (2) block days, once a full cycle passes (A-days and B-days are finished), Joseph is charged $25$ cents on top of the original $25$ cents. Because the test-taker did not account for this, the answer would have been overestimated. - $$14.75$ || The student did not account for block days. Since there are two block days, once a full cycle passes (A-days and B-days are finished), a student is charged $25$ cents on top of the beginning amount of $25$ cents. A student may have made another additional mistake. The student did not account for the initial 25 cents. This value is underestimated yet overestimated at the same time. - $$7.75$ || While the student may have correctly divided the number of weekdays in half to get the following answer, he or she may have forgotten to subtract one from her index and multiply it to the difference. Make sure to start with $$0.25$ on the sequence instead of $$0.50$ cents. + $$7.50$ || The student who answered this correctly deserves a round of applause for paying attention and being careful. Never be too hasty, despite the timer hanging over your head. Always make sure to double check the question and answer before making a hasty jump towards your pseudo-answer. In case the individual guessed, here is an explanation. Since there two different block days (A- and B-block), the number of school days in which the student missed the return of the pencil is greater than the per cycled block day. Therefore, if Joseph borrowed a pencil in B-day but forgot to pay back, he will have to give $25$ cents for forgetting to return the pencil and $25$ more cents per cycled block days in which Joseph forgot to return the pencil (i.e B-day). As such, an individual will have to pay for the $30$ missed days as opposed to the $60$ found in the problem. Finally, since an individual must pay $25$ cents, the student must be charged first $25$ cents, not $0$. Along with using the correct formula, the individual can finally finish this problem. - $$7.25$ || Since there two different block days (A- and B-block), the number of school days in which the student missed the return of the pencil is greater than the per cycled block day. If Joseph borrowed a pencil in B-day but forgot to pay back, he will have to give $25$ cents and $25$ more cents per cycled block days in which Joseph forgot to return the pencil (i.e B-day). As such, an individual starts off with $$0.25$ in debt. If the individual did not make this mistake, be sure not to simply multiply $$0.25$ by the number of days.

{The following arithmetic sequence is listed: $\{-\frac{1}{3},0,\frac{1}{3},\frac{2}{3},1,1\frac{1}{3},1\frac{2}{3},2,\dots \}$ Which of the following is the formula that represents the sequence directly using any index $i$ to find term $x_i$? |type="()"} - $x_i=\frac{1}{3}-\frac{1}{3}(i+1)$ || The formula used is incorrect for three reasons: [1.] the sequence starts at $-\frac{1}{3}$, not $\frac{1}{3}$; [2.] the sequence adds $\frac{1}{3}$, not $-\frac{1}{3}$; finally, [3.] the index term does not start at $-1$ (i.e. notice $i+1$ instead of $i-1$). - $x_i=\frac{1}{3}-\frac{1}{3}(i-1)$ || The formula used is incorrect for two reasons: [1.] the sequence starts at $-\frac{1}{3}$, not $\frac{1}{3}$; [2.] the sequence adds $\frac{1}{3}$, not $-\frac{1}{3}$. - $x_i=-\frac{1}{3}+\frac{1}{3}i$ || The formula used is incorrect because the index term does not start at $0$ (i.e. notice $i$ instead of $i-1$). + $x_i=-\frac{1}{3}+\frac{1}{3}(i-1)$ || The student is able to successfully make an arithmetic formula when asked. - $x_i=-\frac{1}{3}+\frac{1}{3}(i+1)$ || The formula used is incorrect because the index term does not start at $-1$ (i.e. notice $i+1$ instead of $i-1$).

{If the first term is $x$ in a sequence, the last $501$^st term of the sequence is $9,000$, and the constant difference is $20$, what is the second term of the sequence? |type="()"} - $-1,000$ || Be sure not to rush! Notice that the question asked what is the second term, not the first term. Since $a_{501}=x+20(501-1)=x+10000=9000$, the first term $x=-1,000$. However, to get the second term, one must add 20 to the answer found. + $-980$ || Goob job! For any individual that guessed, notice that the question asked what is the second term, not the term before the first one. Since $a_{501}=x+20(501-1)=x+10000=9000$, the first term $x=-1,000$. However, to get the second term, one must add 20 to the answer found. - $-1,020$ || Be sure not to subtract your answer! Notice that the question asked what is the second term, not the term before the first one. Since $a_{501}=x+20(501-1)=x+10000=9000$, the first term $x=-1,000$. However, to get the second term, one must add 20 to the answer found. - $1,000$ || Take a look again at the Algebra! Since $a_{501}=x+20(501-1)=x+10000=9000$, the first term $x=-1,000$. However, to get the second term, one must add 20 to the answer found. - $980$ || Take a look again at the Algebra! Since $a_{501}=x+20(501-1)=x+10000=9000$, the first term $x=-1,000$. However, to get the second term, one must add 20 to the answer found.

{Given $a_i$ is the term of a sequence at index $i$, the first term of a sequence is $x$, and the constant difference is $d$, which of the following must be TRUE? Indicate all such statements. |type="[]"} + $a_1=x;a_i=x+d(i-2)+d$ || Note that a sequence can still follow using a recursive formula. As such, knowing that $a_i=x+d(i-1)$, $a_{i-1}=x+d([i-1]-1)=x+d(i-2)$. In which case, $a_1=x;a_i=a_{i-1}+d$ which is equivalent to $a_1=x;a_i=x+d(i-2)+d$. - $\frac{x-a_i}{d}-1=i$ || Note that the normal arithmetic sequence is $a_i=x+20(i-1)$. If one wants to find index $i$, then subtract $x$, divide by $20$, and $1$ to yield $\frac{a_i-x}{20}+1=i$. + $a_i-x+d=di$ || Note that the normal arithmetic sequence is $a_i=x+20(i-1)$. Since you multiply the constant difference $20$ to $i-1$, distribute it to each term in the parenthesis. Given $a_i=x+20(i-1)=x+20i-20$, try to isolate $20i$ on the right side by subtracting $x$ and adding $20$: $a_i-x+20=20i$. </quiz>

Geometric Sequence

As always, if you are unable to understand, try a few examples of numbers to think of in your head. Let the common ratio $r=2$ and let $x_1=1.5$. The next term $x_2=x_{1}r$, so $x_2=1.5\cdot 2=3$. If you keep the pattern going for each term of the sequence, you would get the following: $\{1.5,3,6,12,24,48,\dots \}$

Recursive Formula

As with the arithmetic formula, you can find the recursive formula and the direct formula for a geometric sequence. Since every term is multiplied by common ratio $r$, let any term of index $i$ be represented by $x_i$. To find the next term requires knowing the previous term. Ergo, $x_i=r{x}_{i-1}$

Direct Formula

As with the arithmetic sequence, Let us chart the recursive geometric formula. $x_1=a;x_k=r{x}_{k-1}$.

$$\begin{array}{cc}k& x\\ 1& a\\ 2& ar\\ 3& a{r}^2\\ 4& a{r}^3\\ ⋮& ⋮\end{array}$$

If you think about it, the table above is basically an exponential function, $f(k)=a{b}^k$, starting at $k=1$. Write out the function as $x=a{r}^k$. We get near our answer. Our independent variable $k$ is translated to the right $1$ unit, so $x=a{r}^{k-1}$ is our function. In fact, we found our direct relationship. Rewrite it the way we normally write it and we found our direct formula: $x_k=a{r}^{k-1}$.

Template:ExampleRobox To find the answer, we need to know the common ratio $r$ of the sequence above. Pick any arbitrary term in the sequence and apply it to the direct geometric formula: $x_2=a{r}^{2-1}$. Find $r$:

$11.2=22.4r.$

$r=0.5=\frac{1}{2}.$

Knowing the value of $r$, you can find the $18$^th term in the sequence by using the direct formula:

$x_{18}=22.4{\left(\frac{1}{2}\right)}^{18-1}$

$x_{18}=22.4(0.00000762939)$

$$x_{18}=0.00017089843$$Template:Robox/Close

As always, these examples are things you can work through yourself or follow along so that you can see how to do a problem.

Series

It is great to find these patterns to these sequences, but is that the only use we have for these sequences? As always, the answer in math is never no in regards to a low amount of utility (i.e. usefulness). A function does not only describe the pattern associated with numbers but also predicts the graph created when plotted using $x$ for its inputs and $y$ for its outputs. Using the terms of a sequence, can we determine a sum? We can, and it is called a series.

There is always a way to find the sum of a sequence by force: add up the terms one by one and get the answer. Mathematicians are lazy people and don't want to do more work than necessary. That is to say, mathematicians work smart, not hard.

Arithmetic Series

Template:ExampleRobox This is a classic example among many mathematicians. You may be reaching for your calculator, but that will definitely take a long time to punch in all those numbers, and you may accidently make mistakes while writing out the expression. The question is, how do we solve this in the short amount of time we are given on the CLEP exam? Notice that there is a pattern at play: the first term of the series, $1$, plus the last term of the series, $100$, will give us $101$. The second term of the series, $2$, plus the second-to-last term of the series, $99$, will give us $101$. In fact, for all $100$ terms in the series, each pair of numbers, according to their "placement," adds to $101$. Since there are $50$ pairs in this series, and the sum of each pair of numbers always yields $101$,

$$101\cdot 50=5050\text{ is the sum of the expression above.}$$Template:Robox/Close

Let us make a conjecture for the above statement. After all, we are not robots that are simply fed a method and do something the same way. We will contract parts of the above expression in the question above to see if our method will be useful for any number of terms. Let $n$ be the number of terms in the arithmetic series $A_E$, let $S$ be the sum of $A_E$, and let $S_m$ be the column in which the same method we used in Example 2.1.1.a will, hopefully, get us the sum $S$.

$$\begin{array}{cccc}n& A_E& S& S_m\\ 2& 1+2& 3& (1+2)\cdot 1=3\\ 3& 1+2+3& 6& (1+3)\cdot 1.5=6\\ 4& 1+2+3+4& 10& (1+4)\cdot 2=10\\ 5& 1+2+3+4+5& 15& (1+5)\cdot 2.5=15\\ ⋮& ⋮& ⋮& ⋮\\ n& 1+2+\dots +(n-1)+n& S& (1+n)\cdot \frac{n}{2}=S\end{array}$$

Notice how the odd terms have this weird (or perhaps odd) behavior where not all pairs of terms have the same sum (because there is not an even amount of pairs). While it may seem that using the method in Example 2.1.1.a will work for all odd number of terms based on the table above, maybe it is not true for some really high odd term. Therefore, it is important to prove this is true. For now, let us simply define the formula to be true for both even and odd number of terms.

A finite set of things is something that has a determined number of anythng within that set. An infinite set is a set not described by a determined number of the amount within that set.

Note: the above formula is not known as the "Gaussian method" among mathematicians. This wikibooks will simply refer to this method as the "Gaussian method."^{[see footnote 1.]}

Template:ExampleRobox This seems like an impossible problem to someone who does not know about the Gaussian method or the properties of an arithmetic sequence. However, since you have paid attention, you can figure out for yourself. For the purposes of giving an example, this Wikibooks will explain. We do not know how many terms there are in the sequence, so let's find out how many terms there are. Since
$$S_A=\frac{a+x_k}{2}\cdot k,$$

$$600=\frac{50+250}{2}\cdot k.$$

This subject is not called College Algebra for nothing, so do some algebra. $$600=\frac{300}{2}\cdot k$$

$$600=150k$$

$$40=k$$

Note, however, that we are not looking for the number of terms in the series. We want to know the constant difference of the arithmetic series. Therefore, use the arithmetic sequence direct formula: $x_k=a+d(k-1)$! $$x_{40}=250$$

$$250=50+d(40-1)$$

$$200=39d$$

$$\frac{200}{39}=d$$

$$5\frac{5}{39}=d$$Template:Robox/Close

Example 2.2.1.b helps us know how to find the number of terms and the constant difference. Sometimes, you may simply not know one or both of the information you needed to find in Example 2.2.1.b., more often the number of terms. The next example helps illustrate the usefulness of knowing about the properties of arithmetic series and arithmetic sequences.

Template:ExampleRobox To find the sum, we need to know how many terms there are; otherwise we cannot use the Gaussian method. Because we do not know the sum of the series, let's use the direct arithmetic sequence formula (since we know which terms correspond to place). We want to find the number of terms, $k$, so use the last term ($x_k=2,504$).

First, find the constant difference. To get from $4$ to $9$, we need to add $5$ to $4$. Therefore, the constant difference is $d=5$. To put it more formally: if recursive formula $x_1=4;x_2=4+d$ and $x_2=9$, then $x_1=4;9=4+d$ means $d=5$.

Second, find the number of terms. If $x_k=2,504$, $d=5$, and $x_1=a=4$, then $2,504=4+5(k-1)$. Solve for $k$:

$$2,500=5(k-1)$$

$$500=k-1$$

$$501=k.$$

Finally, use the Gaussian Method:

$$S_A=\frac{4+2,504}{2}\cdot 501$$

$$S_A=\frac{2,508}{2}\cdot 501$$

$$S_A=1,254\cdot 501$$

$$S_A=628,254.$$Template:Robox/Close

As you can see, an arithmetic series can be useful in describing any type of sequence. Now that we have sufficiently explored arithmetic series, let's prove that the Gaussian method works for odd number of terms.

Note: what you are about to learn is NOT required for the curriculum. If you do not understand this proof, do not worry, for it does not matter for the CLEP exam. These proofs are only to build a mathematical understanding of the concepts. As such, you may skip these if you want.

CLEP Practice Problems: Check Your Understanding

<quiz display=simple> {Given the arithmetic series $50+45+30+\dots -700$, determine the number of terms in the arithmetic series. |type="{}"} { 151 _5 }. (Note: typing a comma will give you a decimal, so do not type a comma.) ||To find the number of terms in the arithmetic series,

{Given the arithmetic series $50+45+30+\dots -700$, determine the sum of the arithmetic series. |type="{}"} { -49075|-49075 _8}. (Note: typing a comma will give you a decimal, so do not type a comma.)

{Function $S$ has the following properties: $S(x)=x\text{ if }-20\le x\le 50$ and $S(x)=3x+1\text{ if }50<x\le 200$. A student decided to add all terms for all $x\in \mathbb{Z}$ or for all x that belong to the set of integers. What is the sum of the terms of function $S$ for all $x\in \mathbb{Z}$? |type="()"} - $56,625$ || You were so close! You forgot to add $1,065$ to the sum. For the individual who guessed, know that this question was difficult. First, given that $S(x)=x\text{ if }-20\le x\le 50$, since adding all $x$ for this piece of the function is equivalent to $-20-19-18-\dots +48+49+50$, $\frac{-20+50}{2}k$ is the sum of that piece. To find the number of terms in that sequence, use the direct arithmetic formula: knowing $x_k=50$, $50=-20+(k-1)$. Solve for $k$ to get $71=k$. Now find the sum: $\frac{-20+50}{2}\cdot (71)=1,065$. Do the same process for the other piece. Simply substitute those values of $x$ to that piece: $3(51)+1=154$ and $3(200)+1=601$. Notice that the constant difference is $3$ for every term, because $3(52)+1=157$, and $157-154=3$. Since we do not know how many terms there are, solve for $k$: $601=154+3(k-1)⟹447=3(k-1)⟹149=k-1⟹150=k$. Find the sum: $\frac{154+601}{2}\cdot (150)=56,625$. Finally, add the two sums of each piece: $56,625+1,065=57,690$. - $57,641$ || You got some part of it right, but some other part of it wrong. While you were able to determine the sum for all $-20\le x\le 50$, you mistakenly used $50\le x\le 200$ instead of $50<x\le 200$. For the individual who guessed, know that this question was difficult. First, given that $S(x)=x\text{ if }-20\le x\le 50$, since adding all $x$ for this piece of the function is equivalent to $-20-19-18-\dots +48+49+50$, $\frac{-20+50}{2}k$ is the sum of that piece. To find the number of terms in that sequence, use the direct arithmetic formula: knowing $x_k=50$, $50=-20+(k-1)$. Solve for $k$ to get $71=k$. Now find the sum: $\frac{-20+50}{2}\cdot (71)=1,065$. Do the same process for the other piece. Simply substitute those values of $x$ to that piece: $3(51)+1=154$ and $3(200)+1=601$. Notice that the constant difference is $3$ for every term, because $3(52)+1=157$, and $157-154=3$. Since we do not know how many terms there are, solve for $k$: $601=154+3(k-1)⟹447=3(k-1)⟹149=k-1⟹150=k$. Find the sum: $\frac{154+601}{2}\cdot (150)=56,625$. Finally, add the two sums of each piece: $56,625+1,065=57,690$. + $57,690$ || The person who got this right deserves a round of applause because this question is trickier than you thought at first glance. For the individual that guessed, know that this question is a little more difficult than you most likely expected. First, given that $S(x)=x\text{ if }-20\le x\le 50$, since adding all $x$ for this piece of the function is equivalent to $-20-19-18-\dots +48+49+50$, $\frac{-20+50}{2}k$ is the sum of that piece. To find the number of terms in that sequence, use the direct arithmetic formula: knowing $x_k=50$, $50=-20+(k-1)$. Solve for $k$ to get $71=k$. Now find the sum: $\frac{-20+50}{2}\cdot (71)=1,065$. Do the same process for the other piece. Simply substitute those values of $x$ to that piece: $3(51)+1=154$ and $3(200)+1=601$. Notice that the constant difference is $3$ for every term, because $3(52)+1=157$, and $157-154=3$. Since we do not know how many terms there are, solve for $k$: $601=154+3(k-1)⟹447=3(k-1)⟹149=k-1⟹150=k$. Find the sum: $\frac{154+601}{2}\cdot (150)=56,625$. Finally, add the two sums of each piece: $56,625+1,065=57,690$. - $57,450$ || Note that your answer is very wrong. You may have used the correct numbers for $S(x)=x\text{ if }-20\le x\le 50$, but the individual determined that there were 70 terms instead of 71 (make sure to include the first term). Second, make sure to use the correct numbers for $S(x)=3x+1\text{ if }50<x\le 200$. Use all numbers above 50. For the individual that guessed, know that this question is a little more difficult than you most likely expected. First, given that $S(x)=x\text{ if }-20\le x\le 50$, since adding all $x$ for this piece of the function is equivalent to $-20-19-18-\dots +48+49+50$, $\frac{-20+50}{2}k$ is the sum of that piece. To find the number of terms in that sequence, use the direct arithmetic formula: knowing $x_k=50$, $50=-20+(k-1)$. Solve for $k$ to get $71=k$. Now find the sum: $\frac{-20+50}{2}\cdot (71)=1,065$. Do the same process for the other piece. Simply substitute those values of $x$ to that piece: $3(51)+1=154$ and $3(200)+1=601$. Notice that the constant difference is $3$ for every term, because $3(52)+1=157$, and $157-154=3$. Since we do not know how many terms there are, solve for $k$: $601=154+3(k-1)⟹447=3(k-1)⟹149=k-1⟹150=k$. Find the sum: $\frac{154+601}{2}\cdot (150)=56,625$. Finally, add the two sums of each piece: $56,625+1,065=57,690$. - $56,776$ || Note that your answer is very wrong. You may have used the correct numbers for $S(x)=x\text{ if }-20\le x\le 50$, but the individual determined that there were 70 terms instead of 71 (make sure to include the first term). Second, make sure to use the correct numbers for $S(x)=3x+1\text{ if }50<x\le 200$. Use all numbers above 50. Finally, if you wanted to find the sum of the two pieces, then you had to add. For the individual that guessed, know that this question is a little more difficult than you most likely expected. First, given that $S(x)=x\text{ if }-20\le x\le 50$, since adding all $x$ for this piece of the function is equivalent to $-20-19-18-\dots +48+49+50$, $\frac{-20+50}{2}k$ is the sum of that piece. To find the number of terms in that sequence, use the direct arithmetic formula: knowing $x_k=50$, $50=-20+(k-1)$. Solve for $k$ to get $71=k$. Now find the sum: $\frac{-20+50}{2}\cdot (71)=1,065$. Do the same process for the other piece. Simply substitute those values of $x$ to that piece: $3(51)+1=154$ and $3(200)+1=601$. Notice that the constant difference is $3$ for every term, because $3(52)+1=157$, and $157-154=3$. Since we do not know how many terms there are, solve for $k$: $601=154+3(k-1)⟹447=3(k-1)⟹149=k-1⟹150=k$. Find the sum: $\frac{154+601}{2}\cdot (150)=56,625$. Finally, add the two sums of each piece: $56,625+1,065=57,690$.

{The first term of an arithmetic sequence is $a<0$. Let there be two separate sequences, $\left\{x_i\right\}=a+d(i-1)$ and $\left\{a_k\right\}=a+d(k-1)$, with $500$ terms in each sequence. If the constant difference is $d<0$ for $\left\{x_i\right\}$, but the constant difference is $d>0$ for $\left\{a_k\right\}$, which of the following must be true? Indicate all such statements |type="[]"} + $x_1-x_2-\dots -x_{499}-x_{500}-a_1-\dots -a_{499}-a_{500}=998a$ + $x_1+x_2+\dots +x_{499}+x_{500}+a_1+\dots +a_{499}+a_{500}=-1,000a$ + $x_{500}<0$ </quiz>

Note, this practice quiz is unfinished.

Proof of Arithmetic Series

Note: what you are about to learn is NOT required for the curriculum. If you do not understand this proof, do not worry, for it does not matter for the CLEP exam. These proofs are only to build a mathematical understanding of the concepts. As such, you may skip these if you want.

Geometric Series

While it is amazing one can find the sum of an arithmetic sequence, what about the sum of a geometric sequence? Thankfully, mathematicians have found a way to calculate this idea. This theorem (not conjecture, as you will hopefully see with the geometric series proof) is one of many ways mathematicians have found solutions to problems coming from the abstract. As always, it is best to understand the way you can use a formula first before getting the formula introduced. To invoke deeper learning, we want to understand the concept, not reciprocate the formula outloud on a test and say, "I know the answer."

Template:ExampleRobox Let's define a sum $S$ for the geometric sequence: $1+2+4+8+\dots +512=S$

Notice that all the terms except $x_1=1$ has a factor $2$. Therefore, move the first term to the other side by subtracting $1$: $2+4+\dots +512=S-1$. Then, factor the two on the left side: $$2(1+2+4+\dots +256)=S-1$$

The key discovery of this exercise is noticing that $1+2+4+\dots +256$ is $S$ but without $512$, so given that $S=1+2+4+8+\dots +512$,

$$1+2+4+\dots +256=S-512$$

Solve using elementary algebra:

$$2S-1024=S-1$$

$$2S-S=-1+1024$$

$$S=1023$$Template:Robox/Close

Notice how we solved the problem. We said that the sum must equal something positive, so we determined that if it does equal something, we could figure it out in some way without having to do any long calculations. Can we use this method for some general geometric sequence? This how we determine a formula. In your Exploration, you will be tasked with presenting a proof for the formula of the geometric series. We will still give you the formula, but you must prove it yourself in the next exploration. Of course, we will also present another way to prove the geometric series formula in the next section.

Proof of Geometric Series

Note: what you are about to learn is NOT required for the curriculum. If you do not understand these concepts, do not worry, for it does not matter for the CLEP exam. These proofs are only to build a mathematical understanding of these concepts. As such, you may skip these if you want.

Sigma Notation

How many of you are tired of writing out the expressions? How many of you are tired of figuring out how many expressions there are in a series? Where is this short-hand that we conveniently have for an arithmetic or geometric sequence? Thankfully, all of your answers are coming soon.

Say you want to write out the expression $\underset{500}{\underbrace{1+1+1+1+\dots +1+1}}$, in which there are $500$ terms in the series, each of which are the number $1$. We hopefully can evaluate the series without having to think too much about it, or even think about a short-hand about this expression immediately without too much thought. However, the purpose of this easy exercise is to introduce a new notation.

Let $\sum$ denote the sum of a series. Located at the bottom of the symbol is the start of the series at index $i=a$ (blue) and the top of the symbol is the last index of the series, $k$ (red). The term inside the parentheses represents the sequence that the series follows, $f(i)$ (orange). The use of the notation of is shown below $$\sum _{i=a}^k\left(f\left(i\right)\right).$$Here, we know that $\{1,1,1,1,\dots 1,1\}$ must have the sequence $x_i=1+0(i-1)=1=f\left(i\right)$.

Interpreting the meaning of $f(i)$ tells us that for any number in the sequence, the $i$^th term of the expression is $1$. Knowing that the sequence starts at index $1$, meaning $i=1$, and there are $500$ terms, meaning $k=500$, we may write the sigma notation as below: $$\sum _{i=1}^{500}\left(1\right).$$

It is important to write the sigma representation having $f(i)$ in parenthesis. Often, if you don't have it, you can confuse other terms by accident. It is for this reason that this Wikibooks recommends writing the $f(i)$ term in parentheses.

Template:Calculus/Def Some more ways in which sigma notation is written is often a short-hand, especially when handwriting. This should be used if you want to save time.

• $\sum _{i=a}^k(f(i))=\sum _{a\le i\le k}(f(i))$
• $\sum _{i=a}^{\mathrm{\infty }}(f(i))=\sum _{i\ge a}(f(i))$

Before we jump into our exploration, it may be a good idea to introduce some more expressions that can be rewritten into its sigma form equivalents. Some of the expressions may not even follow from a formula we have seen so far. Nevertheless, let us continue with this concept. Template:ExampleRobox A good idea before getting started is to look for any possible patterns. Because you are working with sigma notation, check whether there has to be a pattern involving addition.

Look at the numerator of the first two terms. Let us assume we are starting at index $i=1$. To get from the first term in the numerator to the second term in the numerator, one simply has to add 1 to the first term. Similarly, to get from the first term in the denominator to the second term in the denominator, one simply has to add 2 to the first term. Both of these are true on a term-by-term basis.

$f(i)=\frac{i+1}{i+2}$. This is easy to verify for every term in its respective index.

Because this function is true, let us use this to find the number of terms in the expression (i.e., look at the final term in the expression).

Let $f(k)=\frac{71}{72}=\frac{k+1}{k+2}$. There are many ways to solve for $k$, the last term of the expression, in this equation. We will show the standard way.

$\frac{k+1}{k+2}=\frac{71}{72}$

$\iff 72\cdot (k+1)=71\cdot (k+2)$

$\iff 72k+72=71k+142$

$\iff k=70$

We have demonstrated this expression involves 70 terms. With the index, number of terms, and function ready, we may now write the sigma equivalent: $$\sum _{i=1}^{70}\left(\frac{i+1}{i+2}\right)◼$$ Alternatively, one can change the index to get a completely different sigma notation. Notice how when the index starts at an $i>1$, the formula and the final term's index also changes. Here, we will look specifically at $i=2$. Keeping in mind the index represents the placement along the series, going from 2 to 3 is simply a matter of changing along with the numerator. Because the denominator is simply the same idea except displaced by 1 more than the index, the denominator follows a pattern of $i+1$. Therefore, the function of the sigma notation is

$g(i)=\frac{i}{i+1}$. This is easy to verify for every term in its respective index.

Use this function to find the the final term's index.

Let $f(k)=\frac{71}{72}=\frac{k}{k+1}$. We will solve this one the simplest way. Because both the numerator and the denominator are constant, non-zero terms, and $k$ is simply an expression of the index for its final term, one can simply set the numerator equal to the other numerator, and the same is true for the denominator. That is,

$\{\begin{array}{c}k=71\\ k+1=72\end{array}$

This is perhaps the easiest systems of equations one had to deal with this year learning college algebra. From this, it is obvious $k=71$.

Knowing the index starting point, function, and final index, the sigma form of this situation is $$\sum _{i=2}^{71}\left(\frac{i}{i+1}\right)◼$$ Notice how the pattern changed. Quite intriguing, would you not say? Template:Robox/Close Some expressions are too hard to determine simply from looking. These are likely going to involve factors multiplied to another. THe best way to determine such a pattern would be to divide terms to see what the factors are. Template:ExampleRobox There seems to be no pattern here. The best thing to do is to determine any similarities between terms and see if there seems to be a pattern from there.

Notice how the first and fourth term are zero. This means that there is a term that allows us to obtain a value of zero. Let us assume they are linear functions. If $k=0$ is the index of the first term and $k=3$ is the third term, then a possible summation function is $\sum _{k\ge 0}(k(k-3))$. Let us apply this and see if it works.

• $k=0\Rightarrow 0$
• $k=1\Rightarrow -2$
• $k=2\Rightarrow -2$
• $k=3\Rightarrow 0$
• $k=4\Rightarrow 4$
• $k=5\Rightarrow 10$

It seems this function works for each term. Therefore, this works.

$\sum _{k=0}^{\mathrm{\infty }}(k(k-3))=\sum _{k=0}^{\mathrm{\infty }}(k^2-3k)◼$

Template:Robox/Close More examples will be added later. The next exploration will ask you write out a variety of expressions using sigma form: Template:Question-answer

Sigma Simplification Techniques

In some textbooks, this section would be called rules. We call them what they really are: techniques to simplify common sigma representations.

As always, if you are confused, write out some terms. The above sigma representation states that for any constant term $c$ added $k$ times, the resulting sum is equivalent to $\underset{k}{\underbrace{c+c+c+\dots +c+c+c}}=ck$. Never forget that multiplication is repeated addition. This axiom you heard in elementary school is still important to this day.

This second "rule" can be simplified as the following. Since $\sum _{i=1}^k\left(ci\right)$, we can conclude that $\sum _{i=1}^k\left(ci\right)=1c+2c+3c+\dots +c(k-2)+c(k-1)+ck$. Each term in the expression $1c+2c+3c+\dots +c(k-2)+c(k-1)+ck$ has common factor $c$, so $c(1+2+3+\dots +(k-2)+(k-1)+k)$. Notice that each term on the inside of the parenthesis is a basic series where $i$ starting at $1$ goes to final term $k$ as a sum, so

$$\sum _{i=1}^k\left(ci\right)=c\sum _{i=1}^k\left(i\right).$$

This one is harder to see why it is true. As always, however, try it out by hand. Remember that sigma notations are short hands of sums that have some sort of formula, so write out a term in case you are ever confused. Notice that $\sum _{i=1}^k(c+i)=(c+1)+(c+2)+(c+3)+(c+4)+\dots +(c+k-1)+(c+k)$. Because adding is commutative, and because the parentheses do not change the sum of the series, you may group terms such that $(c+1)+(c+2)+(c+3)+(c+4)+\dots +(c+k-1)+(c+k)=\underset{k}{\underbrace{c+c+c+c+\dots +c+c}}+(1+2+3+\dots +k-1+k)$ We have already determined the sigma notation for each grouping, so we may put it together to say the following is true:

$$\sum _{i=1}^k(c+i)=kc+\sum _{i=1}^k\left(i\right).$$

Realize that the above "rule" is simply an extension of the previous one. If $f(i)=i$, then $\sum _{i=1}^k(c+f\left(i\right))=kc+\sum _{i=1}^k\left(f\left(i\right)\right)$. Realize that any function can comply with this rule. By extension:

$$\sum _{i=1}^k\left(cf\left(i\right)\right)=c\sum _{i=1}^k\left(f\left(i\right)\right)$$

Using this general knowledge, one can also argue the next following "rule":

Given the way we found the same identities, you may do the proof for this one by yourself as an exercise in sigma notation. The best way to understand this new notation is to practice it. (Of course, this Wikibooks will provide plenty of practice.)

Along with that practice, you may also try to show that the identity for the next "rule" is also true:

With this, you now have a sufficient foundation for the necessary tools needed to prove and disprove statements as well as create your own identity. Before we move on the next section, we must mention the sigma identity for each of the series type we learned in this wikibooks.

Sigma Identities for Series Types

What is the sigma identity of an arithmetic series? Before we give you the identity, it is important to understand arithmetic series. If the person reading this wikibooks has jumped from one section to the other, we may recommend you read anything from the previous if the current reasoning seems to not make sense. First, how do we write an arithmetic sequence using a formula? Like this:
$$x_k=a+(k-1)d.$$
The formula for how to find an arithmetic series is this:
$$S_a=\frac{a+x_k}{2}k.$$
Notice how $x_k$ is in both formulas, so substitute $x_k$ for $a+(k-1)d$ and you get:
$$S_a=\frac{2a+(k-1)d}{2}k=ak+\frac{(k-1)d}{2}k=[2a+(k-1)d]\frac{k}{2}.$$
Because $x_k=a+(k-1)d.$ is also a formula that helps us find each term in the series, we have finally created our sigma identity:

We can use the same process to find the sum of a geometric series. You may consider this your next exercise proving that the sigma form of the geometric series is the same one shown below.

Check Your Understanding

The problems below are basically comprehension questions. If you can do all the problems below, you understand all you need to know about sigma notations. Keep in mind, these problems are much harder than usual CLEP problems; however, being able to do these problems with proficiency proves understanding of the material at a deeper level, which means you are better prepared for easier problems and harder problems down the line. Template:Question-answer

Infinite Geometric Series

There is this old and famous paradox that perplexed many who listened to it. The problem stems from the Greek thinker Zeno of Elea. This old adage was the first example of reductio ad absurdum (disproving a statement by showing that the application of the statement will lead to a contradiction, and so the original statement cannot be true). However, today, a version is no longer a paradox, and it took the invention of calculus to prove this enigma is not an enigma^{[see footnote 3]}.

The version of the problem that demonstrates the enigma is proven and shown below.

Note: the paradox will be shown soon.

The age-old questions of most students in mathematics arise: "When are we going to need this?" The next exploration gives one problem in which everything you have learned will be tested. This is definitely harder than many CLEP exam questions; however, problems are what make you better at math. You learn as you do, and so you must.

Proof of Infinite Geometric Series

$\sum _{i=1}^{\mathrm{\infty }}\left(a{r}^{i-1}\right)=\frac{a}{1-r}\text{ if }0<\left|r\right|<1$. How do we prove this is true? Again, this is way above most curriculum required textbooks. However, the goal of many mathematics textbooks is to get the user inspired in math and to use those concepts into other fields of interest. No matter what given field a person is going into, they are going to need to solve problems: a history major needs to understand what a person meant; an English major needs to find the right words that help them convey their ideas; a Science major will have to apply mathematical concepts to communicate ideas about science. As such, learn to love problems, for conquering a problem will make you stronger.

Note: what you are about to learn is NOT required for the curriculum. If you do not understand these concepts, do not worry, for it does not matter for the CLEP exam. These proofs are only to build a mathematical understanding of these concepts. As such, you may skip these if you want.

Prerequisite for the proof: Imagine we want to find values of some function $f(x)$. Define this function as $f(x)=\frac{x^2-4}{x-2}$. If $x=2$, then $f(2)=\frac{(2{)}^2-4}{(2)-2}=\frac{0}{0}$. Well, it seems this value is undefined when $x=2$, so it seems it does not equal anything. However, let's try graphing this rational function.

First, let us divide: $\frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2$. It seems once we simplify the rational expression, we are left with $x+2$. However, This is not exactly true because our function is still undefined at $x=2$, so we must add a constraint: $f(x)=x+2\text{ if }x\ne 2$. To put it more formally:

$$f(x)=\{\begin{array}{ccc}x+2,& \text{if}& -\mathrm{\infty }<x<2\\ \frac{x^2-4}{x-2},& \text{if}& x=2\\ x+2,& \text{if}& 2<x<\mathrm{\infty }\end{array}$$

If we graph the relation, we find that there is a "hole" at that point in the function. Let's use values close to 2 so that we abide by those rules.

$$\begin{array}{cc}x& f(x)\\ 1.9& (1.9)+2=3.9\\ 1.99& (1.99)+2=3.99\\ 1.999& (1.999)+2=3.999\\ 2& \text{undefined}\\ 2.001& (2.001)+2=4.001\\ 2.01& (2.01)+2=4.01\\ 2.1& (2.1)+2=4.1\end{array}$$

It becomes clear that as values get "closer and closer" to $2$, $f(2)$ "approaches" $4$. What is "close"? How close are your eyes to the screen? We would most likely say, "it is 45 cm. away from the screen." In which case, let's use the distance (i.e. absolute value) to define the word "close": $|2.001-2|=0.001$ and $|1.999-2|=0.001$. (Answer this question: why did we use absolute values?)

Let's define this process using two variables $x_1$ and $x$: If $0<|x_1-x|<n$, then those two values are "close" or "approaches $L$ of function $f(x)$," where $n$ can be any value you want it to be, as long as it tolerates our definition. We can conclude the following: $|f(x)-L|$ is "close" when $|x_1-x|$ is close, given that $0<|f(x)-L|<n_f$. If we think about it, we have sufficiently proven what "close" means. Let us put this definition^{[see footnote 4]} up front:

For any $0<|f(x)-L|<n_f$, there exists a $0<|x_1-x|<n$.

Now, as mentioned, mathematicians like to "work smarter, not harder," so a mathematician does not want to write out "as $x$ approaches $2$, $f(x)$ approaches $4$." Instead, we write out using our fancy notation: $\underset{x\to 2}{\lim }f(x)=4$. This $\lim$ is called "limit." All this notation says is that "as we limit our values of $x$ approaching ($\to$) $2$, we find that $f(x)$ approaches $4$." Now, you have a sufficient foundation for limits and are ready to begin the proof.

Footnotes

1. The WikiBooks simply refers the formula used as the "Gaussian method" because of a famous mathematical tale. There was this intelligent mathematician named Carl Friedrich Gauss who, in second grade, was asked to find the sum of $1+2+3+4+\dots +99+100$. The teacher that assigned this problem simply wanted some peace and quiet. However, Gauss was able to come up the answer using the same method in Example 1.3.1.a. Legend says that he did not write the method and did all the calculations in his head; he handed the piece of paper with just the answer. Nevertheless, the teacher had to check to see if he was right. It turns out, he was indeed right. If the reader needs to refer to this formula to other mathematicians, simply call it the "sum of an arithmetic series formula," although it does not have the same ring to it as the previous name.
2. It is very common for many textbooks to use the following formula for a geometric series: $S_g=\frac{a(1-r^k)}{1-r}$. Both are technically correct; however, this Wikibooks has decided to use $S_g=\frac{a(r^k-1)}{r-1}$ because it is easier to remember that you subtract the first term from the final term at index $k$.
3. Although described technically correct, calculus was merely one piece of the puzzle that helped in solving the issue. The version that is described is shown in the example, but the original problem did involve calculus concepts that will not be talked about in this Wikibooks since it does not help us understand infinite geometric series.
4. Note the formal definition is technically complete; however, the variables refered to the distances are a little different. Instead of $0<|x_1-x|<n$, write it as $0<|x_1-x|<\delta$ (read $\delta$ aloud as "delta"). Instead of $0<|f(x)-L|<n_f$, write it as $0<|f(x)-L|<ϵ$ (read $ϵ$ aloud as "epsilon"). Consequently, the definition given for "close" changes: "For any $0<|f(x)-L|<ϵ$, there exists a $0<|x_1-x|<\delta$." Also note that the definition may be written a little different as "for any $ϵ>0$, there is a $\delta >0$ such that $|f(x)-L|<ϵ$ and $|x_1-x|<\delta$." For the interest of not confusing the reader, we decided to make the definition more simple to understand.
5. Note that the limit can be used for any function. Let $L(x)=2x$. Say we want to find $\underset{x\to 10}{\lim }L(x)$. We can do that because as we get closer and closer to $10$, we find that $L(x)$ gets closer and closer to $20$. However, we can simplify the process by saying that $\underset{x\to L}{\lim }f(x)=f(L)$, given that a function is continuous. Because there are no jumps or asymptotes in $f(i)=\left(a{r}^{i-1}\right)$ as long as $0\le |r|<1$, we can state $\sum _{i=1}^{\mathrm{\infty }}f(i)=\frac{a}{1-r}\text{ if }0\le \left|r\right|<1$. This is why we can use the limit and say that $\underset{k\to \mathrm{\infty }}{\lim }\sum _{i=1}^k\left(a{r}^{i-1}\right)=\sum _{i=1}^{\mathrm{\infty }}\left(a{r}^{i-1}\right)=\frac{a}{1-r}\text{ if }0\le \left|r\right|<1$.