\(\)Consider

$$
f(x)=(a+x)^r\label{eq:axr}
$$

Recall the MacLaurin Series

$$
\begin{align}
f(x)&=\sum _{k=0}^{\infty }{\frac {f^{(k)}(0)}{k!}}\,x^{k}\nonumber
\end{align}
\nonumber
$$

The \(k\)^{th} derivative of equation \(\eqref{eq:axr}\)

$$
f^{(k)}(x) = r\,(r-1)\cdots(r-k+1)(a+x)^{r-k}
$$

Substitute \(x=0\) to find the derivatives at \(0\)

$$
f^{(k)}(0)=r\,(r-1)\cdots(r-k+1)\,a^{r-k}
$$

Apply the MacLaurin Series to equation \(\eqref{eq:axr}\)

$$
\begin{align}
f(x)=(a+x)^r&=\sum _{k=0}^{\infty }\frac{r\,(r-1)\cdots(r-k+1)}{k!}\,a^{r-k}\,x^k
\end{align}
$$

Isaac Newton generalized binomial theorem for \(r\in\mathbb{C}\)

$$
\shaded{
(a+x)^r=\sum_{k=0}^{\infty}{r \choose k}\,a^{r-k}\,x^k,\quad\text{where }{r \choose k}=\frac{r\,(r-1)\cdots(r-k+1)}{k!}
}
\label{eq:newton}
$$

The *binomial coefficient* \({r \choose k}\)

$$
\begin{align}
\frac{(r)_k}{k!}&=\frac{r(r-1)(r-2)\cdots (r-k+1)}{k(k-1)(k-2)\cdots1} \nonumber \\
&= \prod _{i=1}^{k}\frac{(r-(i-1))}{i} = \prod _{i=0}^{k-1}\frac{r-i}{i}
\end{align}
$$

The series converges for \(r\geq0\land r\in\mathbb{N}\), or for \(|x|\lt|a|\)

$$
\begin{align}
(a+x)^r&=\sum_{k=0}^{\infty}{r \choose k}\,a^{r-k}\,x^k\nonumber\\
&=a^r+r\,a^{r-1}\,x+\frac{r(r-1)}{2!}\,a^{r-2}\,x^2+\frac{r(r-1)(r-2)}{3!}\,a^{r-3}\,x^3+\cdots
\end{align}
$$

##
Binomial series

Consider equation \(\eqref{eq:newton}\) for \(a=1\), gives the *Binomial series*

$$
\shaded{
(1+x)^r=\sum_{k=0}^{\infty}{r \choose k}\,x^k,\quad\text{where }{r \choose k}=\frac{r\,(r-1)\cdots(r-k+1)}{k!}
}
$$

This series converges when

- \(|x|\lt1\), converges absolutely for any complex number \(r\).
- \(|x|\gt1\), converges only when \(r\) is a non-negative integer, what makes the series finite.

##
Special cases

**1**) where \(a=1\), converges for \(|x|\lt1\)

$$
\begin{align}
(1+x)^{r} &= \sum_{k=0}^{\infty}\frac{(r)_k}{k!}\,x^k \nonumber \\
&= 1+r\,x+\frac{r(r-1)}{2!}\,x^2+\frac{r(r-1)(r-2)}{3!}\,x^3+\cdots
\end{align}
$$

**2**) the

*negative binomial series*, converges for \(|x|\lt1\)

Apply the Negated Upper Index of Binomial Coefficient identity \({r \choose k}=(-1)^k{k-r-1 \choose k}\)

$$
\begin{align}
(a+x)^r&=\sum_{k=0}^{\infty}{r \choose k}\,a^{r-k}\,x^k\nonumber\\
&=\sum_{k=0}^{\infty}{k-r-1 \choose k}(-1)^k\,\,a^{r-k}\,x^k
\end{align}
$$

Substitute \(x\to -x\) and \(m\to -m\)

$$
\begin{align}
(a-x)^{-r}&=\sum_{k=0}^{\infty}{k+r-1 \choose k}(-1)^k\,\,a^{-r-k}\,(-x)^k\nonumber\\
&=\sum_{k=0}^{\infty}{k+r-1 \choose k}\cancel{(-1)^k}\,\,a^{-r-k}\,\cancel{(-1)^k}\,x^k\nonumber\\
&=\sum_{k=0}^{\infty}{k+r-1 \choose k}\,a^{-r-k}\,x^k
\end{align}
$$

For \(a=1\)

$$
\begin{align}
(1-x)^{-r}&=\sum_{k=0}^{\infty}{k+r-1 \choose k}\,x^k\nonumber\\
\end{align}
$$