# Algebra/Binomial Theorem

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

$n! = n \times (n-1) \times (n-2) \times (n-3) \times (n-4) \times...... 3 \times 2 \times 1$

0 factorial is equal to 1.

$0! = 1$

Proof of 0 factorial = 1

$n!= n \times (n-1)!$
When n = 1,
$1!= 1 \times (1-1)!$
$1=1 \times 0!$
And thus,
$0!=1$

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 $x \ne 0$ set z = y / x

$(x + y)^n = x^n (1 + z)^n$ .

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

$(1 + z)^n = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} z^k$ .

Noting that

$(x + y)^n = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} x^{n - k} y^k$

is symmetric in x and y, the identity

$\begin{pmatrix} n \\ n - k \end{pmatrix} = \begin{pmatrix} n \\ k \end{pmatrix}$

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

A recursive relationship between the $\begin{pmatrix} n \\ k \end{pmatrix}$ may be established by considering

$(1 + z)^{n+1} = (1 + z)(1 + z)^n = \sum_{k=0}^{n + 1} \begin{pmatrix} n + 1 \\ k \end{pmatrix} z^k = (1 + z) \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} z^k$

or

$\sum_{k=0}^{n + 1} \begin{pmatrix} n + 1 \\ k \end{pmatrix} z^k = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} z^k + \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} z^{k + 1} = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} z^k + \sum_{k=1}^{n + 1} \begin{pmatrix} n \\ k - 1 \end{pmatrix} z^k$ .

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

$\begin{pmatrix} n + 1 \\ k \end{pmatrix} = \begin{pmatrix} n \\ k \end{pmatrix} + \begin{pmatrix} n \\ k - 1 \end{pmatrix}$

for k ranging from 1 through n, and

$\begin{pmatrix} n + 1 \\ n + 1 \end{pmatrix} = \begin{pmatrix} n \\ n \end{pmatrix} = \frac{n!}{(n-n)!n!} = \frac{n!}{n!} = 1$
$\begin{pmatrix} n + 1 \\ 0 \end{pmatrix} = \begin{pmatrix} n \\ 0 \end{pmatrix} = \frac{n!}{(n-0)!0!} = \frac{n!}{n!} = 1$ .

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

Show

$\begin{pmatrix} n \\ k \end{pmatrix} = \frac{n!}{k!(n - k)!}$

(proof by induction on n).

A useful identity results by setting $z = 1$

$\sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} = 2^n$ .

## The visual way to do the binomial theorem

(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.