# Binomial Series

## Key Questions

• Binomial Series

${\left(1 + x\right)}^{\alpha} = {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\alpha \\ n\end{matrix}\right) {x}^{n}$,

where ((alpha),(n))={alpha(alpha-1)(alpha-2)cdot cdots cdot(alpha-n+1)}/{n!}.

Let us look at this example below.

$\frac{1}{\sqrt{1 + x}}$

by rewriting a bit,

$= {\left(1 + x\right)}^{- \frac{1}{2}}$

by Binomial Series,

$= {\sum}_{n = 0}^{\infty} \left(\begin{matrix}- \frac{1}{2} \\ n\end{matrix}\right) {x}^{n}$

by writing out the binomial coefficients,

=sum_{n=0}^infty{(-1/2)(-3/2)(-5/2)cdots(-{2n-1}/2)}/{n!}x^n

by simplifying the coefficients a bit,

=sum_{n=0}^infty(-1)^n{1cdot3cdot5cdot cdots cdot(2n-1)}/{2^n n!}x^n

I hope that this was helpful.

• Pascal's triangle gives the binomial coefficients.

Pascal's Triangle

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
...

Binomial Coefficients

$\left(\begin{matrix}0 \\ 0\end{matrix}\right)$

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

$\left(\begin{matrix}2 \\ 0\end{matrix}\right)$ $\left(\begin{matrix}2 \\ 1\end{matrix}\right)$ $\left(\begin{matrix}2 \\ 2\end{matrix}\right)$

$\left(\begin{matrix}3 \\ 0\end{matrix}\right)$ $\left(\begin{matrix}3 \\ 1\end{matrix}\right)$ $\left(\begin{matrix}3 \\ 2\end{matrix}\right)$ $\left(\begin{matrix}3 \\ 3\end{matrix}\right)$

$\left(\begin{matrix}4 \\ 0\end{matrix}\right)$ $\left(\begin{matrix}4 \\ 1\end{matrix}\right)$ $\left(\begin{matrix}4 \\ 2\end{matrix}\right)$ $\left(\begin{matrix}4 \\ 3\end{matrix}\right)$ $\left(\begin{matrix}4 \\ 4\end{matrix}\right)$
...

I hope that this was helpful.

• To understand Ismail's answer, it is worth recalling some notations:

((n),(k))=(n!)/((n-k)!k!), where $n , k \in \mathbb{N}$

n! =n.(n-1)...2.1