Find coefficient of $x^3$ in (2+x) ^(3/2)/(1-x)

Question

I can expand $\dfrac{(2+x)^{3/2}}{1-x}=(1+x+x^2+\ldots)\left({3/2\choose0}+{3/2\choose1}(x+1)+{3/2\choose2}(x+1)^2+\ldots\right)$, but that doesn't seem to lead anywhere.

Answer 1

Your idea is right but it helps to factor out the $2$ in the numerator so the binomial expansion is in $x$ instead of $x+1$:

\begin{eqnarray*} \dfrac{(2+x)^{3/2}}{1-x} &=& \dfrac{2^{3/2}(1+\frac{x}{2})^{3/2}}{1-x} \\ &=& 2^{3/2}(1+x+x^2+\ldots)\left({3/2\choose0}+{3/2\choose1}\frac{x}{2}+{3/2\choose2}\left(\frac{x}{2}\right)^2+{3/2\choose3}\left(\frac{x}{2}\right)^3+\ldots\right). \end{eqnarray*}

Multiplying out and gathering the $x^3$ terms we get the $x^3$ coefficient:

\begin{eqnarray*} x^3\mbox{ coeff.} &=& 2^{3/2}\left(1+\frac{3}{2}\cdot\frac{1}{2} + \dfrac{\frac{3}{2}\cdot\frac{1}{2}}{2} \cdot\frac{1}{2^2} + \dfrac{\frac{3}{2}\cdot\frac{1}{2}\cdot\frac{-1}{2}}{2\cdot3} \cdot\frac{1}{2^3}\right) \\ &=& \dfrac{2^{3/2}}{128}\left(128+96+12-1\right) \\ && \\ &=& \dfrac{235\sqrt{2}}{64} \end{eqnarray*}