<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] When dealing with a sum like you are here (summing $n$ terms for some general expression), I would almost always recommend that you use $\Sigma$-notation, for it tidies up a lot of the algebraic mess you have to deal with in your induction proof. With that in mind, you may write your claim as follows. 
Claim: For any $n\geq 1$, the statement
$$
S(n) : \sum_{i=1}^n\frac{1}{i(i+2)}=\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}
$$
is true.
Base step ($n=1$): $S(1)$ says that $\sum_{i=1}^1\frac{1}{i(i+2)}=\frac{1}{3}=\frac{3}{4}-\frac{5}{12}$, and this is true.
Induction step ($S(k)\to S(k+1)$): Fix some $k\geq 1$, and assume that
$$
S(k) : \sum_{i=1}^k\frac{1}{i(i+2)}=\frac{3}{4}-\frac{2k+3}{2(k+1)(k+2)}
$$
is true. To be proved is that
$$
S(k+1) : \sum_{i=1}^{k+1}\frac{1}{i(i+2)}=\frac{3}{4}-\frac{2k+5}{2(k+2)(k+3)}
$$
follows. Beginning with the left side of $S(k+1)$,
\begin{align}
\sum_{i=1}^{k+1}\frac{1}{i(i+2)}&= \sum_{i=1}^k\frac{1}{i(i+2)}+\frac{1}{(k+1)(k+3)}\tag{by defn.}\\[1em]
&= \left(\frac{3}{4}-\frac{2k+3}{2(k+1)(k+2)}\right)+\frac{1}{(k+1)(k+3)}\tag{by $S(k)$}\\[1em]
&= \frac{3}{4}-\frac{2k^2+7k+5}{2(k+1)(k+2)(k+3)}\tag{common denom.}\\[1em]
&= \frac{3}{4}-\frac{(2k+5)(k+1)}{2(k+1)(k+2)(k+3)}\tag{factor}\\[1em]
&= \frac{3}{4}-\frac{2k+5}{2(k+2)(k+3)},\tag{simplify}
\end{align}
one arrives at the right side of $S(k+1)$, thereby showing $S(k+1)$ is also true, completing the inductive step. 
By mathematical induction, the claim $S(n)$ is true for all $n\geq 1$. $\blacksquare$
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