Show that the sums are equal to: a) If $\sum_{k=1}^∞ \frac{1}{k^2} = \frac{π^2}{6}$ show that  $\sum_{k=1}^∞ \frac{1}{(2k+1)^2} = \frac{π^2}{8}$
b) Show that $\sum_{k=1}^∞ \frac{1}{k(k+1)(k+2)} = \frac{1}{4}$
Hint for b): Find A, B, C so that for ∀k:
$\frac{1}{k(k+1)(k+2)}$= $\frac{A}{k}$ +$\frac{B}{k+1}$+ $\frac{C}{k+2}$
I managed to do a) but don't know how to do b)
a) $\sum_{k=1}^∞ \frac{1}{k^2}$ -  $\sum_{k=1}^∞ \frac{1}{(2k)^2}$ =   $\sum_{k=1}^∞ \frac{1}{(2k+1)^2}$
$\sum_{k=1}^∞ \frac{1}{k^2}$ -  $\sum_{k=1}^∞ \frac{1}{4}\frac{1}{k^2}$ =  $\frac{π^2}{6}$ - $\frac{1}{4}$$\frac{π^2}{6}$=$\frac{π^2}{8}$

If you find the constants $A,B,C$, they are $\frac{1}{2},-1,\frac{1}{2}$ respectively, then you term
$\frac{1}{k(k+1)(k+2)}=\frac{1}{2k}-\frac{1}{k+1}+\frac{1}{2(k+2)}=\frac{1}{2}(\frac{1}{k}-\frac{1}{k+1})+\frac{1}{2}(\frac{1}{k+2}-\frac{1}{k+1})$ if you take the series (Harmonic series) you limit is $\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$.
Recall that $\sum_{1}^n (a_{k+1}-a_{k})=a_{n+1}-a_1$ then $\sum_{K=1}^{\infty}(a_{k+1}-a_{k})=\lim_{n\rightarrow \infty}\sum_{1}^n (a_{k+1}-a_{k})=\lim_{n\rightarrow \infty}a_{n+1}-a_1$
