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 $n\geq1$?

Let $$p(n):\displaystyle\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$$
Put $n=1\;,$ We get $$\displaystyle \frac{1}{1\cdot 3} = \frac{3}{4}-\frac{5}{2\cdot 2\cdot 3} = \frac{4}{12}$$
So it is true for $n=1$
Now Put $n=k\;,$ We get $$\displaystyle \frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{k\cdot(k+2)}=\frac{3}{4} - \frac{(2k+3)}{2(k+1)(k+2)}$$
Now Using $p(k)\;,$ We will prove for $p(k+1)$
So $$\displaystyle p(k+1):\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{(k+1)\cdot(k+3)}=\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{k\cdot(k+2)}+\frac{1}{(k+1)\cdot (k+3)}$$ 
So $$\displaystyle p(k+1) = \frac{3}{4} - \frac{(2k+3)}{2(k+1)(k+2)}+\frac{1}{(k+1)(k+3)}= \frac{3}{4}-\frac{1}{(k+1)}\left\{\frac{2k+3}{2(k+2)}-\frac{1}{(k+3)}\right\}$$
$$\displaystyle = \frac{3}{4}-\frac{1}{(k+1)}\left\{\frac{2k^2+9k+9-2k-4}{2(k+2)(k+3)}\right\} = \frac{3}{4}-\frac{1}{2(k+1)(k+2)(k+3)}\cdot (2k+5)\cdot (k+1)=\frac{3}{4}-\frac{(2k+5)}{2(k+2)(k+3)}$$
So $p(k)$ We have prove for $p(k+1).$
