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

Jul 12, 2017

${\left(1 + 2 x\right)}^{5} = 1 + 10 x + 40 {x}^{2} + 80 {x}^{3} + 80 {x}^{4} + 32 {x}^{5}$

#### Explanation:

Recall the Binomial Theorem

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

Where:

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

is the combinatorial, which are also the numbers from the $n t h$ row of Pascal's Triangle

Thus we have:

${\left(1 + 2 x\right)}^{5} = {\sum}_{r = 0}^{5} \setminus \left(\begin{matrix}5 \\ r\end{matrix}\right) \setminus {\left(1\right)}^{5 - r} \setminus {\left(2 x\right)}^{r}$
$\text{ } = {\sum}_{r = 0}^{5} \setminus \left(\begin{matrix}5 \\ r\end{matrix}\right) \setminus {\left(2 x\right)}^{r}$

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

$\text{ } = \left(1\right) + \left(5\right) \left(2 x\right) + \left(10\right) \left(4 {x}^{2}\right) + \left(10\right) \left(8 {x}^{3}\right) + \left(5\right) \left(16 {x}^{4}\right) + \left(1\right) \left(32 {x}^{5}\right)$

$\text{ } = 1 + 10 x + 40 {x}^{2} + 80 {x}^{3} + 80 {x}^{4} + 32 {x}^{5}$