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?
 A: If induction is not mandatory,
let $p(m)=\dfrac{am+b}{(2m-1)(2m+1)}$
$$p(k)-p(k+1)=\dfrac{ak+b}{(2k-1)(2k+1)}-\dfrac{a(k+1)+b}{(2k+1)(2k+3)}$$
$$=\dfrac{ak+4b+a}{(2k-1)(2k+1)(2k+3)}$$
We need $a=0,4b+a=1\iff b=?$
So in essence, we have a Telescoping series
A: When $n=1$,
$$\begin {align}
\sum_{k=1}^n\frac 1{(2k-1)(2k+1)(2k+3)}&=\frac 1{(2.1-1)(2.1+1)(2.1+3)}\\
&=\frac 1{15}\\
&=\frac{1(1+2)}{3(2.1+1)(2.1+3)}\\
&=\frac{n(n+2)}{3(2n+1)(2n+3)}
\end {align}$$
Assume the result to be true for $n=m$ 
We show it is also true for $n=m+1$,
$$\begin {align}
\sum_{k=1}^{m+1}\frac 1{(2k-1)(2k+1)(2k+3)}&=\sum_{k=1}^{m}\frac 1{(2k-1)(2k+1)(2k+3)}\\
&+\frac 1{(2(m+1)-1)(2(m+1)+1)(2(m+1)+3)}\\
&=\frac {m(m+2)}{3(2m+1)(2m+3)}+\frac{1}{(2m+1)(2m+3)(2m+5)}\\
&=\frac {1}{(2m+1)(2m+3)}\times\bigg(\frac{m(m+2)}{3}+\frac 1{2m+5}\bigg)\\
&=\frac {1}{(2m+1)(2m+3)}\times\bigg(\frac{(m^2+2m)(2m+5)+3}{3(2m+5)}\bigg)\\
&=\frac {1}{(2m+1)(2m+3)}\times\bigg(\frac{2m^3+9m^2+10m+3}{3(2m+5)}\bigg)\\
&=\frac {1}{(2m+1)(2m+3)}\times\frac{(2m+1)(m+1)(m+3)}{3(2m+5)}\\
&=\frac{(m+1)(m+3)}{3(2m+5)(2m+3)}\\
&=\frac{\big(m+1\big)\big((m+1)+2\big)}{3\big(2(m+1)+1\big)\big(2(m+1)+3\big)}\\
\end {align}$$
Hence,
by Principle of Mathematical Induction, the result holds for all $n\geq1$.
A: Since $k$ is a dummy variable, you should induct on $n$. In other words, show the claim is right when $n=1$, then show $$\sum_{k=1}^p\tfrac{1}{(2k-1)(2k+1)(2k+3)}=\tfrac{p(p+2)}{3(2p+1)(2p+3)}\implies\sum_{k=1}^{p+1}\tfrac{1}{(2k-1)(2k+1)(2k+3)}=\tfrac{(p+1)(p+3)}{3(2p+3)(2p+5)}.$$
