Proving $\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$ by induction for $n\geq 1$ I'm having an issue solving this problem using induction. If possible, could someone add in a very brief explanation of how they did it so it's easier for me to understand? 
$$\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$$
How do I prove the above equation for all integers where $n\geq1$?

First, check the formula for $n=1$. So:
$$\frac1{1\cdot 3}=\frac34-\frac{2\cdot1+3}{2(1+1)(2+1)}$$
Since this is true, we have shown the so called base case.
Now substitute in the formula $n$ by $n+1$ to get the statement that you have to show:
$$\frac1{1\cdot 3}+\frac1{2\cdot 4}+\cdots+\frac1{(n+1)(n+3)}=\frac34-\frac{2n+5}{2(n+2)(n+3)}\qquad (*)$$
The good news is that you can (and should) assume that the formula is valid for the $n$ first positive integers, so
$$\begin{align}&\left(\frac1{1\cdot 3}+\frac1{2\cdot 4}+\cdots+\frac1{n(n+2)}\right)+\frac1{(n+1)(n+3)}\\
&=\frac34-\frac{2n+3}{2(n+1)(n+2)}+\frac1{(n+1)(n+3)}
\end{align}$$
and you should obtain $(*)$ with straightforward computings.
Perhaps my explanation is not very brief but I hope it to be useful.
