Question

Find the expansion of$\mathbf{(}\mathit{x}\mathbf{+}\mathit{y}{\mathbf{)}}^{\mathbf{6}}$.

Short answer

$x^6+6x^{5}y+15x^4{y}^2+20x^3{y}^3+15x^2{y}^4+6x{y}^5+y^6$

Step-by-step solution

Formula for Binomial theorem

Binomial theorem is:$\mathbf{(}\mathit{x}\mathbf{+}\mathit{y}{\mathbf{)}}^{\mathbf{n}}\mathbf{=}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathbf{ }\left(\begin{array}{l}n\\ j\end{array}\right){\mathit{x}}^{\mathbf{n}\mathbf{-}\mathbf{j}}{\mathit{y}}^{\mathbf{j}}$

Definition combination (order is not important):$\mathit{C}\mathbf{(}\mathit{n}\mathbf{,}\mathit{r}\mathbf{)}\mathbf{=}\left(\begin{array}{l}n\\ r\end{array}\right)\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{r}\mathbf{!}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{!}}$

with$\mathit{n}\mathbf{!}\mathbf{=}\mathit{n}\mathbf{-}\mathbf{(}\mathit{n}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{\dots }\mathbf{-}\mathbf{2}\mathbf{-}\mathbf{1}$

Use Binomial theorem and simplify

The given function is$(x+y{)}^6$ . To find the expansion$x^6{y}^0+x^5{y}^1+x^4{y}^2+x^3{y}^3+x^2{y}^4+x^1{y}^5+x^0{y}^6$ we start with your first variable.

Place the combined power on it, then count the power down to 0 .

Take your second variable and count up from 0 .

Any variable with power 0 is 1 .

$\frac{6!}{6!(6-6)!}\left(x^6\right)+\frac{6!}{5!(6-5)!}\left(x^{5}y\right)+\frac{6!}{4!(6-4)!}\left(x^4{y}^2\right)+\frac{6!}{3!(6-3)!}\left(x^3{y}^3\right)+\frac{6!}{2!(6-2)!}\left(x^2{y}^4\right)+\frac{6!}{1!(6-1)!}\left(x{y}^5\right)+\frac{6!}{0!(6-0)!}\left(y^5\right)$

To find the coefficients, start with the first power and count down, that is 6 choose 6 , 6 choose 5,6 choose 4,6 choose 3,6 choose 2,6 choose 1,6 choose 0.

Use the combination formula and solve: $x^6+6x^{5}y+15x^4{y}^2+20x^3{y}^3+15x^2{y}^4+6x{y}^5+y^6$.