<s>[INST] 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 $n\geq1$?
 [/INST] No induction is necessary: it's a matter of a telescoping sum, if you write:
$$\frac1{k(k+2)}=\frac12\Bigl(\frac 1k-\frac1{k+2}\Bigr)$$
Apply this decomposition to the above sum:
\begin{align*}
\sum_{k=1}^n\frac1{k(k+2)}&=\frac12\sum_{k=1}^n\Bigl(\frac 1k-\frac1{k+2}\Bigr)\\
&=\frac12\Bigl(1\color{red}{-\frac13}+\frac12\color{red}{-\frac14+\frac13-\frac15+\dots+\frac1{n-1}}-\frac1{n+1}\color{red}{+\frac 1n}-\frac1{n+2}\Bigr)\\
&=\frac12\Bigl(1+\frac12-\frac1{n+1}-\frac1{n+2}\Bigr)=\frac34-\frac{n+2+n+1}{2(n+1)(n+2)}.
\end{align*}
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