# What is the binomial theorem?

Feb 12, 2016

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 ${\left(2 x + 3 y\right)}^{5}$.

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

$= {\text{^5C_0(2x)^(5-0)(3y)^0+""^5C_1(2x)^(5-1)(3y)^1+""^5C_2(2x)^(5-2)(3y)^2+""^5C_3(2x)^(5-3)(3y)^3+""^5C_4(2x)^(5-4)(3y)^4+}}^{5} {C}_{5} {\left(2 x\right)}^{5 - 5} {\left(3 y\right)}^{5}$

$= \left(1\right) 32 {x}^{5} \left(1\right) + \left(5\right) \left(16 {x}^{4}\right) \left(3 y\right) + \left(10\right) \left(8 {x}^{3}\right) \left(9 {y}^{2}\right) + \left(10\right) \left(4 {x}^{2}\right) \left(27 {y}^{3}\right) + \left(5\right) \left(2 x\right) \left(81 {y}^{4}\right) + \left(1\right) \left(1\right) \left(243 {y}^{5}\right)$

$= 32 {x}^{5} + 240 {x}^{4} y + 720 {x}^{3} {y}^{2} + 1080 {x}^{2} {y}^{3} + 810 x {y}^{4} + 243 {y}^{5}$.

Feb 16, 2016

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

Instead of doing ${\text{_n"C}}_{r}$ in front of each term, we can use the coefficients in the $n + 1$ row of Pascal's triangle.

In the case of ${\left(2 x + 3 y\right)}^{5}$, the ${\text{_n"C}}_{r}$ 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 ${\left(2 x + 3 y\right)}^{5}$, we get

$1 {\left(2 x\right)}^{5} {\left(3 y\right)}^{0} + 5 {\left(2 x\right)}^{4} {\left(3 y\right)}^{1} + 10 {\left(2 x\right)}^{3} {\left(3 y\right)}^{2} + 10 {\left(2 x\right)}^{2} {\left(3 y\right)}^{3} + 5 {\left(2 x\right)}^{1} {\left(3 y\right)}^{4} + 1 {\left(2 x\right)}^{0} {\left(3 y\right)}^{5}$

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

Simplified, this gives us

$32 {x}^{5} + 240 {x}^{4} y + 720 {x}^{3} {y}^{2} + 1080 {x}^{2} {y}^{3} + 810 {x}^{4} + 243 {y}^{5}$