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 $n\geq1$?
 A: 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.
A: 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*}
A: Hint:
$$\frac { 1 }{ 2 } \left( \frac { 1 }{ n } -\frac { 1 }{ n+2 }  \right) =\frac { 1 }{ n\left( n+2 \right)  } $$
A: 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 $p(k+1).$
A: 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.
A: 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 $n\geq 1$. $\blacksquare$
