First, here's the answer I want to check.

$\displaystyle \frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1} {(2n-1)(2n+1)}=\frac{n}{2n+1}$

$\displaystyle s_1=\frac{1}{3} (check)$

$\displaystyle \frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1} {(2n-1)(2n+1)}+\frac{1}{(2k+1)(2k+3)}$

$\displaystyle =P_k+\frac{1}{(2k+1)(2k+3)}$

$\displaystyle =\frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}$

$\displaystyle =\frac{k+1}{2k+1}$

~~~~~~~~~~~~~~~~~~~

Second, on the problem:

$\displaystyle 2+7+...+(5n-3)=\frac{n}{2}(5n-1)$

I get to:

$\displaystyle P_k+(5k+2)$

$\displaystyle \frac{k}{2}(5k+1)+(5k+2)$

And once again, I get stuck trying to finish the problem off. I know how to get to this point, I just fail to see how to finish it off

Any and all help (and verification for the first problem) will be greatly appreciated.