Proving Sums or Products by Induction I am currently teaching Extension 1 Mathematics and seem to be coming unstuck on a few induction questions. 
$$\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)}=\frac{n}{2n+1}$$
First step is to prove the equation works for "1" - which it does. The second step is to prove the function works for $k+1$ with my working set out as follows:
$$LHS=\sum_{k=1}^n{\frac{1}{(2k-1)(2k+1)}}+\frac{1}{(2n+1)(2n+3)}$$
$$=\frac{n}{2n+1}+\frac{1}{(2n+1)(2n+3)}$$
$$=\frac{n(2n+3)+1}{(2n+1)(2n+3)}$$
After setting out these steps - my text tells me the answer i'm looking for is $\frac{n+1}{2n+3}$ and I cant seem to find how to get from the final step of my working to the final answer! What am I missing here? (this is the second question I have come unstuck on a step before the final answer)
 A: You're stuck at
$$\frac{n(2n+3)+1}{\color{red}{(2n+1)}\color{green}{(2n+3)}}$$
And want to go to:
$$\frac{n+1}{\color{green}{2n+3}}$$
So, you just have to expend and refactor the numerator and hope that a $(2n+1)$ factor will appear!
$$n(2n+3)+1 = 2n^2+3n+1 = (n+1)(2n+1)$$
Therefore:
$$\frac{n(2n+3)+1}{(2n+1)(2n+3)}=\frac{(n+1)(2n+1)}{(2n+1)(2n+3)}=\frac{n+1}{{2n+3}}$$
A: We have $$\frac{2k^2+3k+1}{(2k+1)(2k+3)}=\frac{(2k+1)(k+1)}{(2k+1)(2k+3)} $$
Note that the line on the right of LHS makes no sense, do not omit summation sign. Also be careful about the choice of notation, $n$ or $k$.
A: Fro the induction step we need to prove that
$$\sum_{k=1}^{n+1}\frac{1}{(2k-1)(2k+1)}=\frac{1}{(2n+1)(2n+3)}+\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)}\stackrel{Ind. Hyp.}=\frac{1}{(2n+1)(2n+3)}+\frac{n}{2n+1}\stackrel{?}=\frac{n+1}{2n+3}$$
that is
$$\frac{1}{(2n+1)(2n+3)}+\frac{n}{2n+1}\stackrel{?}=\frac{n+1}{2n+3}$$
$$\frac{1}{(2n+1)(2n+3)}=\frac{n+1}{2n+3}-\frac{n}{2n+1}$$
$$\frac{1}{(2n+1)(2n+3)}=\frac{(2n+1)(n+1)-n(2n+3)}{(2n+1)(2n+3)}$$
$$\frac{1}{(2n+1)(2n+3)}=\frac{\color{red}{2n^2+3n}+1\color{red}{-2n^2-3n}}{(2n+1)(2n+3)}=\frac{1}{(2n+1)(2n+3)}$$
