Prove with induction $\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)(2k+3)}=\frac{n(n+2)}{3(2n+1)(2n+3)}$ Prove with induction the identity
$\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)(2k+3)}=\frac{n(n+2)}{3(2n+1)(2n+3)}$
How can I solve this problem?
Should i set k= (p+1) and n = (p+1), then try to get the left side equal to the right side?

Since $k$ is a dummy variable, you should induct on $n$. In other words, show the claim is right when $n=1$, then show $$\sum_{k=1}^p\tfrac{1}{(2k-1)(2k+1)(2k+3)}=\tfrac{p(p+2)}{3(2p+1)(2p+3)}\implies\sum_{k=1}^{p+1}\tfrac{1}{(2k-1)(2k+1)(2k+3)}=\tfrac{(p+1)(p+3)}{3(2p+3)(2p+5)}.$$
