# Algebra/Binomial Theorem

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

**0 factorial** is equal to 1.

**Proof of 0 factorial = 1**

- When n = 1,
- And thus,

The binomial thereom gives the coefficients of the polynomial

- .

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

- .

The expansion coefficients of are known as the binomial coefficients, and are denoted

- .

Noting that

is symmetric in *x* and *y*, the identity

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

A recursive relationship between the may be established by considering

or

- .

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

for k ranging from 1 through n, and

- .

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

Show

(proof by induction on *n*).

A useful identity results by setting

- .

## The visual way to do the binomial theorem[edit]

(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}= 1x^{0}= 1 (x+1)^{1}= 1x^{1}+1x^{0}= 1 1 (x+1)^{2}= 1x^{2}+2x^{1}+1x^{0}= 1 2 1 (x+1)^{3}= 1x^{3}+3x^{2}+3x^{1}+1x^{0}= 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} = 1x^{10} + 10x^{9} + 45x^{8} + 120x^{7} + 210x^{6} + 252x^{5} + 210x^{4} + 120x^{3} + 45x^{2} + 10x^{1} + 1x^{0}.