2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ remember you only have expand up to and including $x^3$. So multiply out the result you have and collect terms of $O(x^3)$. $\endgroup$ – Chinny84 Sep 5 '14 at 10:38
  • $\begingroup$ Thanks. I've collected coefficients of $1,x,x^2,x^3$ from the second polynomial and got $\sum_{k\ge0}{3/2\choose k}+\sum_{k\ge1}{3/2\choose k}k+\sum_{k\ge2}{3/2\choose k}{k\choose2}+\sum_{k\ge3}{3/2\choose k}{k\choose3}$, but I can't sum those. Is there any other way? $\endgroup$ – k5f Sep 5 '14 at 10:53
2
$\begingroup$

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*}

$\endgroup$
  • $\begingroup$ That's exactly the kind of answer I was looking for. $\endgroup$ – k5f Sep 5 '14 at 14:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.