Question

What is the binomial theorem?

Answer 1

It is a method one may use to expand a binomial expression raised to a positive integer power as follows :

`(x+y)^n=sum_(r=0)^n""^nC_rx^(n-r)y^r`

The combination notation used is defined as follows :

`""^nC_r=(n!)/((n-r)!r!)`

Example :

Expand `(2x+3y)^5`.

This is a binomial (2 terms) raised to an integer power, so the binomial theorem is valid and may be used as follows :

`(2x+3y)^5=sum_(r=0)^5 ""^5C_r(2x)^(5-r)(3y)^r`

`=""^5C_0(2x)^(5-0)(3y)^0+""^5C_1(2x)^(5-1)(3y)^1+""^5C_2(2x)^(5-2)(3y)^2+""^5C_3(2x)^(5-3)(3y)^3+""^5C_4(2x)^(5-4)(3y)^4+""^5C_5(2x)^(5-5)(3y)^5`

`=(1)32x^5(1)+(5)(16x^4)(3y)+(10)(8x^3)(9y^2)+(10)(4x^2)(27y^3)+(5)(2x)(81y^4)+(1)(1)(243y^5)`

`=32x^5+240x^4y+720x^3y^2+1080x^2y^3+810xy^4+243y^5`.

Answer 2

There is a simpler way of expanding a binomial that uses the binomial theorem but takes a more intuitive approach.

Instead of doing `""_n"C"_r` in front of each term, we can use the coefficients in the `n+1` row of Pascal's triangle.

In the case of `(2x+3y)^5`, the `""_n"C"_r` series will be identical to the `6`th row of Pascal's triangle:

The row we want is

`1,5,10,10,5,1`

In order to deal with exponents, know that the exponent on the first term will start at `5` and work its way down to `0`, and the second term will start with an exponent of `0` and work its way up to `5`.

If there is a negative term they will alternate positive, negative, positive, negative, etc.

For `(2x+3y)^5`, we get

`1(2x)^5(3y)^0+5(2x)^4(3y)^1+10(2x)^3(3y)^2+10(2x)^2(3y)^3+5(2x)^1(3y)^4+1(2x)^0(3y)^5`

Note that anything to the `0` power is `1`.

Simplified, this gives us

`32x^5+240x^4y+720x^3y^2+1080x^2y^3+810x^4+243y^5`