# How do you use the binomial series to expand  (1+x^2)^5?

Mar 7, 2018

$1 + 5 {x}^{2} + 10 {x}^{4} + 10 {x}^{6} + 5 {x}^{8} + {x}^{10}$

#### Explanation:

For a binomial expansion:

${\left(x + y\right)}^{n}$ we have:

$\left(\begin{matrix}n \\ r\end{matrix}\right) {x}^{n - r} {y}^{r}$

${\sum}_{r = 0}^{n} \left(\begin{matrix}n \\ r\end{matrix}\right) {x}^{n - r} {y}^{r}$

Where:

((n),(r))=color(white)(0)^n C_(r)=(n!)/(r!(n-r)!)

Beginning with $r = 0$

$\left(\begin{matrix}5 \\ 0\end{matrix}\right) {1}^{5} {\left({x}^{2}\right)}^{0} + \left(\begin{matrix}5 \\ 1\end{matrix}\right) {1}^{4} {\left({x}^{2}\right)}^{1} + \left(\begin{matrix}5 \\ 2\end{matrix}\right) {1}^{3} {\left({x}^{2}\right)}^{2} + \left(\begin{matrix}5 \\ 3\end{matrix}\right) {1}^{2} {\left({x}^{2}\right)}^{3}$

$\to \left(\begin{matrix}5 \\ 4\end{matrix}\right) {1}^{1} {\left({x}^{2}\right)}^{4} + \left(\begin{matrix}5 \\ 5\end{matrix}\right) {1}^{0} {\left({x}^{2}\right)}^{5}$

Next calculate $\left(\begin{matrix}n \\ r\end{matrix}\right)$ and simplify:

$1 + 5 {x}^{2} + 10 {x}^{4} + 10 {x}^{6} + 5 {x}^{8} + {x}^{10}$

Short cut:

${\textcolor{w h i t e}{0}}^{n} {C}_{r} = {\textcolor{w h i t e}{0}}^{n} {C}_{n - r}$

When one of the terms is $1$, it is unnecessary to write out the powers. This was just done for completeness.