Complete induction: $\sum^n_{i=1}\frac{1}{(2i-1)(2i+1)}=\frac{n}{2n+1}$ I am very confused with complete induction. Because in every task there is something different to do, and I never know what to insert (thats my biggest problem).
Here's the example:
Proof with complete induction. Please please help me, because I have exams coming up (I am just becoming a primary school teacher..)

For $n\in\mathbb{N}$:
$$\sum^n_{i=1}\frac{1}{(2i-1)(2i+1)}=\frac{n}{2n+1}$$


For a full solution, proceed like this:  
$n=1$: $$\sum_{i=1}^1 \frac{1}{(2i-1)(2i+1)} = \frac{1}{(2-1)(2+1)} = \frac{1}{3} = \frac{1}{2 \cdot 1 +1},$$
so it holds for $n=1$.  
Assume next that it holds for some generic $n$. You need to show that then it also holds for $n+1$. As it holds for $n$, you can assume that
$$\sum_{i=1}^n \frac{1}{(2i-1)(2i+1)} = \frac{n}{2n+1}. \quad (1),$$
and want to show that
$$\sum_{i=1}^{n+1} \frac{1}{(2i-1)(2i+1)} = \frac{n+1}{2(n+1)+1}. \quad (2)$$
Then:
$$\begin{align}
\sum_{i=1}^{n+1} \frac{1}{(2i-1)(2i+1} &= \sum_{i=1}^n \frac{1}{(2i-1)(2i+1)} + \frac{1}{(2(n+1)-1)(2(n+1)+1)} \\
& = \frac{n}{2n+1} + \frac{1}{(2n+1)(2n+3)} \quad \text{using (1)} \\
& = \frac{n(2n+3)}{(2n+1)(2n+3)} + \frac{1}{(2n+1)(2n+3)} \\
& = \frac{2n^2 +3n +1}{(2n+1)(2n+3)} \\ 
& = \frac{(n+1)(2n+1)}{(2n+1)(2n+3)} = \frac{n+1}{2(n+1)+1},\\
\end{align}$$
which is (2), and
 was to be shown.
