Algebra/Chapter 10/Binomial Theorem

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{{Template:BOOKTEMPLATE/Page}} Template:Navigation 10.3: The Binomial Theorem Template:Wikipedia

Factorials

The notation ' n! ' is defined as n factorial.

n!=n×(n1)×(n2)×(n3)××3×2×1

0 factorial is equal to 1.

0!=1

Proof of 0 factorial = 1

n!=n×(n1)!
When n = 1,
1!=1×(11)!
1=1×0!
And thus,
0!=1

The Binomial Theorem

The binomial thereom gives the coefficients of the polynomial

(x+y)n .

We may consider without loss of generality the polynomial, of order n, of a single variable z. Assuming x0 set z = y / x

(x+y)n=xn(1+z)n .

The expansion coefficients of (1+z)n are known as the binomial coefficients, and are denoted

(1+z)n=k=0n(nk)zk .

Noting that

(x+y)n=k=0n(nk)xnkyk

is symmetric in x and y, the identity

(nnk)=(nk)

may be shown by replacing k by n - k and reversing the order of summation.

A recursive relationship between the (nk) may be established by considering

(1+z)n+1=(1+z)(1+z)n=k=0n+1(n+1k)zk=(1+z)k=0n(nk)zk

or

k=0n+1(n+1k)zk=k=0n(nk)zk+k=0n(nk)zk+1=k=0n(nk)zk+k=1n+1(nk1)zk .

Since this must hold for all values of z, the coefficients of zk on both sides of the equation must be equal

(n+1k)=(nk)+(nk1)

for k ranging from 1 through n, and

(n+1n+1)=(nn)=n!(nn)!n!=n!n!=1
(n+10)=(n0)=n!(n0)!0!=n!n!=1 .

Pascal's Triangle is a schematic representation of the above recursion relation ...

Show

(nk)=n!k!(nk)!

(proof by induction on n).

A useful identity results by setting z=1

k=0n(nk)=2n .

The visual way to do the binomial theorem

Template:Wikipedia (this section is from difference triangles)

Lets look at the results for (x+1)n where n ranges from 0 to 3.

(x+1)0 =          1x0           =                1         
(x+1)1 =        1x1+1x0         =              1   1
(x+1)2 =      1x2+2x1+1x0       =            1   2   1
(x+1)3 =    1x3+3x2+3x1+1x0     =          1   3   3   1

This new triangle is Pascal’s Triangle.

It follows a counting method different from difference triangles.

The sum of the x-th number in the n-th difference  and 
the (x+1)-th number in the n-th difference yields the
(x+1)-th number in the (n-1)-th difference.

It would take a lot of adding if we were to use the difference triangles in the X-gon to compute (x+1)10. However, using the Pascal’s Triangle which we have derived from it, the task becomes much simpler. Let’s expand Pascal’s Triangle.

(x+1)0                                    1
(x+1)1                                  1   1
(x+1)2                                1   2   1
(x+1)3                             1    3   3    1
(x+1)4                           1    4   6   4    1
(x+1)5                         1   5   10   10   5   1
(x+1)6                      1   6   15   20   15   6   1
(x+1)7                   1   7   21   35   35   21   7    1
(x+1)8                1   8   28   56   70   56   28   8    1
(x+1)9              1   9   36   84   126  126  84   36   9    1
(x+1)10          1   10  45   120  210  252  210  120  45   10    1 

The final line of the triangle tells us that

(x+1)10 = 1x10 + 10x9 + 45x8 + 120x7 + 210x6 + 252x5 + 210x4 + 120x3 + 45x2 + 10x1 + 1x0.

Example Problems

Practice Problems

<quiz display=simple points=1/1> {Problem 1: Find the following without using a pencil and paper:}

{992 |type="{}"} { 9801_7 }

{9982 |type="{}"} { 996004_7 }

{1013 |type="{}"} { 1030301_7 } </quiz>

Problem 2: If 3! * 5! * 7! = n!, what is n?