Binomial Theorem

Binomial Expansion
We can apply differentiation to find a certain term of an expansion.

Example:
Find the coefficient of $x^3$ in $\Large \frac{1}{\sqrt{x}(2+\sqrt{x})^3}$.

Conventional method:
$$\begin{align}\dfrac{1}{\sqrt{x}(2+\sqrt{x})^3}&=\dfrac{1}{8\sqrt{x}(1+\frac{\sqrt x}{2})^3}\\
\dfrac{1}{(1+x)^3}&=\sum_{k=0}^{\infty}\binom{-3}{k}x^k=1-3x+6x^2-10x^3+30x^4-\cdots\\ \dfrac{1}{(1+\frac{\sqrt x}{2})^3}&=1-3\frac{\sqrt{x}}{2}+3\frac{x^2}{2}-\cdots\\ \dfrac{1}{8\sqrt{x}(1+\frac{\sqrt x}{2})^3}&=\frac{1}{8\sqrt{x}}(1-3\frac{\sqrt{x}}{2}+3\frac{x^2}{2}-\cdots)\end{align}$$ coefficient of $x^3$ of the given expression $=\dfrac{1}{8} \cdot \binom{-3}{7}\cdot \dfrac{1}{2^7}\\
=\dfrac{1}{8} \cdot \binom{9}{7}\cdot \dfrac{1}{2^7}\\
=-\dfrac{9}{256}$
[In general, $\binom{-n}{k}=(-1)^k\binom{n+k-1}{k}$]

Differentiation:
$$\begin{align}\frac{1}{1+x}&=\sum(-x)^k\\
-\frac{1}{(1+x)^2}&=-\sum(k+1)(-x)^k\\
\frac{2}{(1+x)^3}&=\sum(k+2)(k+1)(-x)^k\\
\frac{1}{(1+x)^3}&=\frac{1}{2}\sum(k+2)(k+1)(-\frac{\sqrt{x}}{2})^k\\
\frac{1}{8\sqrt{x}(1+x)^3}&=\frac{1}{16\sqrt{x}}\sum(k+2)(k+1)(-\frac{\sqrt{x}}{2})^k\\
&=\ldots+\frac{1}{16}(7+2)(7+1)(-\frac{1}{2})^7x^3+\ldots\end{align}$$ coefficient of $x^3=-\dfrac{9}{256}$