How can we simplify $ \sum _{k=-n}^n k^2 q^k $? Is there any nice expression?
Simplifying $ \sum _{k=-n}^n k^2 q^k $
1
$\begingroup$
summation
1 Answers
1
If $q=1,\sum_{-n\le k\le n}k^2q^k=\sum_{-n\le k\le n}k^2=2\sum_{0\le k\le n}k^2$ as $(-t)^2=t^2$
So, $\sum_{-n\le k\le n}k^2q^k=2\frac{n(n+1)(2n+1)}6=\frac{n(n+1)(2n+1)}3$
If $q\ne 1,$ $ \sum _{k=-n}^n q^k=q^{-n} \frac{q^{2n+1}-1}{q-1}$
Apply derivative wrt $q,$ $\sum _{k=-n}^n kq^{k-1}=\frac{d\left(\frac{q^{n+1}-q^{-n}}{q-1}\right)}{dq}$
Now multiply with $q,$ $\sum _{k=-n}^n kq^{k}=q\frac{d\left(\frac{q^{n+1}-q^{-n}}{q-1}\right)}{dq}$
Apply derivative wrt $q$ and multiply with $q$