Generalized Binomial Expansion

By Taylor's series expansion, the $n^{th}$ coefficient $a_n = \dfrac{f^{(n)} (0)}{n!}$ where $f^{(n)} (x)$ is the $n^{th}$ derivative of $f(x)$ . Therefore, $a_3 = \dfrac{f'''(0)}{3!}$ .

$\begin{aligned} a_3 & = \frac{\left[\frac{d^3}{dx^3}\sqrt{1+x}\right]_{x=0}}{3!} \\ & = \frac{1}{6} \times \frac{d^2}{dx^2} \left[\frac{1}{2} (1+x)^{-\frac{1}{2}} \right]_{x=0} \\ & = \frac{1}{12} \times \frac{d}{dx} \left[-\frac{1}{2} (1+x)^{-\frac{3}{2}} \right]_{x=0} \\ & = \frac{1}{24} \times \left[-\frac{3}{2} (1+x)^{-\frac{5}{2}} \right]_{x=0} \\ & = \frac{3}{48} = \frac{1}{16} = \boxed{0.0625} \end{aligned}$

We can rewrite $\sqrt{1+x}$ as $(1+x)^{1/2}$ . Then, using the generalized binomial theorem ,

$(1 + x)^{1/2} = \sum\limits_{k=0}^{\infty} {1/2 \choose k} 1^{1/2 - k} x^{k}$

Setting $k = 3$ gives

${1/2 \choose 3} = \frac{(\frac{1}{2}) (\frac{1}{2} - 1) (\frac{1}{2} - 2)}{3!} = \frac{\frac{3}{8}}{6} = \frac{1}{16} = 0.0625$