A-level Mathematics/OCR/FP1/Summation of Series

Summation of a Series

In [[../../C2|Core Two]] we learned about arithmetic and geometric progression, but if we need to sum an arithmetic progression over a large range it can become very time consuming. There are formulae that can allow us to calculate the sum. Note that these formulae only work if we start from 1; we will see how to calculate summations from other end points in the example below. The formulae are:

$\sum _{r=1}^{n}1=n$

$\sum _{r=1}^{n}r=\frac{1}{2}n(n+1)$

$\sum _{r=1}^n{r}^2=\frac{1}{6}n(n+1)(2n+1)$

$\sum _{r=1}^n{r}^3=\frac{1}{4}{n}^2{(n+1)}^2$

We also need to know this general result about summation:

$\sum _{r=1}^{n}a{r}^b=a\sum _{r=1}^n{r}^b$

You can see why this is true by thinking of the expanded form:

$(a\times 1^b+a\times 2^b+a\times 3^b+...+a\times n^b)\equiv a(1^b+2^b+3^b+...+n^b)$

Example

Find the sum of the series $\sum _{x=3}^{10}3x^3+4x^2+5x$.

1. First we need to break the summation into its three separate components.

   $\sum _{x=3}^{10}3x^3+\sum _{x=3}^{10}4x^2+\sum _{x=3}^{10}5x$

2. Next we need to make them start from one. We then need to subtract the sum of the numbers not included in the summation.

   $\sum _{x=1}^{10}3x^3-\sum _{x=1}^{2}3x^3+\sum _{x=1}^{10}4x^2-\sum _{x=1}^{2}4x^2+\sum _{x=1}^{10}5x-\sum _{x=1}^{2}5x$

3. Now we use the identities to calculate the individual sums. Remember to include the co-efficients.

   $3[\frac{1}{4}10^2{(10+1)}^2-\frac{1}{4}{2}^2{(2+1)}^2]+4[\frac{1}{6}10(2\times 10+1)(10+1)-\frac{1}{6}2(2\times 2+1)(2+1)]$ $+5[\frac{1}{2}10(10+1)-\frac{1}{2}2(2+1)]$

4. Now we need to perform a lot of arithmetic. This can be done by hand or utilizing a calculator.

   $\frac{3}{4}(100\times 121-4\times 9)+4\frac{1}{6}(10\times 21\times 11-10\times 3)$ $+\frac{5}{2}(10\times 11-2\times 3)$

   $=\frac{3}{4}(12100-36)+\frac{2}{3}(2310-30)+\frac{5}{2}(110-6)$

   $=10828$

5. The sum of the series $\sum _{x=3}^{10}3x^3+4x^2+5x=10828$.