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, show that this is true for $n=1$:
$\sum\limits_{k=1}^{1}\frac{1}{k(k+2)}=\frac34-\frac{2+3}{2(1+1)(1+2)}$
Second, assume that this is true for $n$:
$\sum\limits_{k=1}^{n}\frac{1}{k(k+2)}=\frac34-\frac{2n+3}{2(n+1)(n+2)}$
Third, prove that this is true for $n+1$:
$\sum\limits_{k=1}^{n+1}\frac{1}{k(k+2)}=$
$\color\red{\sum\limits_{k=1}^{n}\frac{1}{k(k+2)}}+\frac{1}{(n+1)(n+3)}=$
$\color\red{\frac34-\frac{2n+3}{2(n+1)(n+2)}}+\frac{1}{(n+1)(n+3)}=$
$\frac34-\frac{2(n+1)+3}{2(n+2)(n+3)}$
Please note that the assumption is used only in the part marked red.
