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

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.

• remember you only have expand up to and including $x^3$. So multiply out the result you have and collect terms of $O(x^3)$. – Chinny84 Sep 5 '14 at 10:38
• 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? – k5f Sep 5 '14 at 10:53

Your idea is right but it helps to factor out the $2$ in the numerator so the binomial expansion is in $x$ instead of $x+1$:
Multiplying out and gathering the $x^3$ terms we get the $x^3$ coefficient: