# How do you use the binomial series to expand  (2+kx)^8 ?

Jan 21, 2017

#### Explanation:

${\left(a + b\right)}^{n} = {\sum}_{r = 0}^{n} \left(\begin{matrix}n \\ r\end{matrix}\right) {a}^{n - r} {b}^{r}$

((n),(r))=(n!)/(r!(n-r)!)

Therefore,

${\left(2 + k x\right)}^{8} = {\sum}_{r = 0}^{8} \left(\begin{matrix}8 \\ r\end{matrix}\right) {2}^{8 - r} {\left(k x\right)}^{r}$

$= \left(\begin{matrix}8 \\ 0\end{matrix}\right) {2}^{8} + \left(\begin{matrix}8 \\ 1\end{matrix}\right) {2}^{7} \left(k x\right) + \left(\begin{matrix}8 \\ 2\end{matrix}\right) {2}^{6} {\left(k x\right)}^{2} + \left(\begin{matrix}8 \\ 3\end{matrix}\right) {2}^{5} {\left(k x\right)}^{3} + \left(\begin{matrix}8 \\ 4\end{matrix}\right) {2}^{4} {\left(k x\right)}^{4} + \left(\begin{matrix}8 \\ 5\end{matrix}\right) {2}^{3} {\left(k x\right)}^{5} + \left(\begin{matrix}8 \\ 6\end{matrix}\right) {2}^{2} {\left(k x\right)}^{6} + \left(\begin{matrix}8 \\ 7\end{matrix}\right) 2 {\left(k x\right)}^{7} + \left(\begin{matrix}8 \\ 8\end{matrix}\right) {2}^{0} {\left(k x\right)}^{8}$

$= 256 + 1024 k x + 1792 {k}^{2} {x}^{2} + 1762 {k}^{3} {x}^{3} + 1120 {k}^{4} {x}^{4} + 448 {k}^{5} {x}^{5} + 112 {k}^{6} {x}^{6} + 14 {k}^{7} {x}^{7} + {k}^{8} {x}^{8}$