Unsure of Where to Start - Prove by Mathematical Induction Prove by induction
$$\sum_{k=1}^n\frac1{(2k-1)(2k+1)}=\frac n{2n+1}$$
for all $n\ge1.$
I'm unsure of where to start with this question
 A: First you must show that the statement holds for the small $n$ i.e. $n=1$. Then you assume that the statement is true for $n$, and use that assumption to demonstrate that it holds for $n+1$. 
So in general for a proposition $P_n$, you are showing that $P_1$ holds and $P_n\implies P_{n+1}$ 
Edit: if you get stuck -
In this case we have $$P_{n}: \sum_{k=1}^n\frac1{(2k-1)(2k+1)}=\frac n{2n+1},\quad n\geq 1$$
We first check $P_1:$
$$LHS = \frac{1}{1\cdot 3}=\frac{1}{3}=RHS$$
So $P_1$ holds. Now we must assume $P_n$ holds, and attempt to prove $P_{n+1}$
$$\sum_{k=1}^{n+1} \frac{1}{(2k-1)(2k+1)} = \underbrace{\sum_{k=1}^{n} \frac{1}{(2k-1)(2k+1)}}_{P_n} + \frac{1}{(2(n+1)-1)(2(n+1)+1)}$$
$$=\frac{n}{2n+1} +\frac{1}{(2n+1)(2n+3)} =\frac{n(2n+3)+1}{(2n+1)(2n+3)}$$
$$= \frac{2n^2+3n+1}{(2n+1)(2n+3)} =\frac{(2n+1)(n+1)}{(2n+1)(2n+3)} =\frac{n+1}{2(n+1)+1}$$
as required, hence $P_{n+1}$ holds - hence $P_n$ holds $\forall n\in\mathbb{N_{n\geq 1}}$ by induction.
A: You start with the base case.  Evaluate both sides for $n=1$ and show that they are equal.  Then you assume it is true for $j$, so $\sum_{k=1}^j\frac1{(2k-1)(2k+1)}=\frac j{2j+1}$ Now you want to prove it is true for $j+1$.  That adds one more term to the sum, so you would write $\sum_{k=1}^{j+1}\frac1{(2k-1)(2k+1)}=\sum_{k=1}^j\frac1{(2k-1)(2k+1)}+\frac 1{(2j+1)(2j+3)}$ and try to show that equals $\frac {j+1}{2(j+1)+1}$
