# How do you use the binomial series to expand  (x+1)^20?

Nov 9, 2016

#### Answer:

${\left(x + 1\right)}^{20} = {\sum}_{k = 0}^{20} \left(\begin{matrix}20 \\ k\end{matrix}\right) {x}^{20 - k}$

#### Explanation:

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

where ((n),(k)) = (n!)/((n-k)! k!)

Putting $a = x$, $b = 1$ and $n = 20$ we get:

${\left(x + 1\right)}^{20} = {\sum}_{k = 0}^{20} \left(\begin{matrix}20 \\ k\end{matrix}\right) {x}^{20 - k}$

$\textcolor{w h i t e}{{\left(x + 1\right)}^{20}} = \left(\begin{matrix}20 \\ 0\end{matrix}\right) {x}^{20} + \left(\begin{matrix}20 \\ 1\end{matrix}\right) {x}^{19} + \ldots + \left(\begin{matrix}20 \\ 20\end{matrix}\right)$

If you just want a single term from this expansion, then it is probably easiest to calculate directly.

For example, the fourth term is:

((20),(3)) x^17 = (20!)/(17! 3!) x^17 = (20xx19xx18)/(3xx2xx1) x^17 = 1140 x^17

If you want to write out every term, then it may be easier to write out Pascal's triangle as far as the $21$st row - hoping that you make no errors on the way. The $21$st row starts $1 , 20 , 190 , 1140 , \ldots$ consisting of the values:

$\left(\begin{matrix}20 \\ 0\end{matrix}\right) , \left(\begin{matrix}20 \\ 1\end{matrix}\right) , \ldots , \left(\begin{matrix}20 \\ 20\end{matrix}\right)$

conveniently providing all the coefficients you want.