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, 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
$$\begin{align}&amp;\left(\frac1{1\cdot 3}+\frac1{2\cdot 4}+\cdots+\frac1{n(n+2)}\right)+\frac1{(n+1)(n+3)}\\
&amp;=\frac34-\frac{2n+3}{2(n+1)(n+2)}+\frac1{(n+1)(n+3)}
\end{align}$$
and you should obtain $(*)$ with straightforward computings.

Perhaps my explanation is not very brief but I hope it to be useful.
Proof: By mathematical induction. 


$P(1)$: $\frac{1}{1\cdot 3} = \frac{1}{3} = \frac{1}{3} = \frac{4}{12} = \frac{9}{12}-\frac{5}{12} = \frac{3}{4}-\frac{5}{12} = \frac{3}{4}-\frac{5}{2\cdot 2 \cdot 3} = \frac{3}{4}-\frac{2 \cdot 1+3}{2(1+1)(1+2)}$. 
$P(n) \Rightarrow P(n+1)$: Assume as the Inductive Hypothesis: 
$$\small \frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+\cdots+\frac{1}{n(n+2)} = \frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}.$$

We need to show $$\small \frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3 \cdot 5}+\cdots+\frac{1}{(n+1)((n+1)+2)} = \frac{3}{4}-\frac{2(n+1)+3}{2((n+1)+1)((n+1)+2)}.$$

Note that the LHS is $$\small \frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+\cdots+\frac{1}{(n+1)((n+1)+2)} = \frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+\cdots+\frac{1}{(n+1)(n+3)}.$$ Notice the two factors in the denominators of each term increase by $1$ as I go to the right. This means I am able to display the term before $\frac{1}{(n+1)(n+3)}$ by subtracting the two factors by $1$.


$$\begin{align}
\mathrm{LHS} &amp;=
\frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+\cdots+\frac{1}{(n+1)(n+3)} \\
&amp;= 
\frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+\cdots+\frac{1}{n(n+2)}+\frac{1}{(n+1)(n+3)} \\
&amp;=
\left( \frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+\cdots+\frac{1}{n(n+2)}\right)+\frac{1}{(n+1)(n+3)} \\
&amp;= 
\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}+\frac{1}{(n+1)(n+3)} \\
&amp;= 
\frac{3}{4}-\frac{(2n+3)(n+3)}{2(n+1)(n+2)(n+3)}+\frac{2(n+2)}{2(n+1)(n+2)(n+3)}  \\
&amp;= 
\frac{3}{4}+\frac{-(2n+3)(n+3)+2(n+2)}{2(n+1)(n+2)(n+3)} \\
&amp;= 
\frac{3}{4}+\frac{-(2n^2+6n+3n+9)+2n+4}{2(n+1)(n+2)(n+3)} \\
&amp;= 
\frac{3}{4}+\frac{-2n^2-9n-9+2n+4}{2(n+1)(n+2)(n+3)} \\
&amp;= 
\frac{3}{4}+\frac{-2n^2-7n-5}{2(n+1)(n+2)(n+3)} \\
&amp;= 
\frac{3}{4}-\frac{2n^2+7n+5}{2(n+1)(n+2)(n+3)} \\
&amp;= 
\frac{3}{4}-\frac{2n^2+2n+5n+5}{2(n+1)(n+2)(n+3)} \\
&amp;= 
\frac{3}{4}-\frac{2n(n+1)+5(n+1)}{2(n+1)(n+2)(n+3)} \\
&amp;= 
\frac{3}{4}-\frac{(n+1)(2n+5)}{2(n+1)(n+2)(n+3)} \\
&amp;= 
\frac{3}{4}-\frac{2n+5}{2(n+2)(n+3)} \\
&amp;= 
\frac{3}{4}-\frac{2n+2+3}{2(n+1+1)(n+1+2)} \\
&amp;= 
\frac{3}{4}-\frac{2(n+1)+3}{2((n+1)+1)((n+1)+2)}. 
\end{align}$$

Therefore, $\frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+\cdots+\frac{1}{n(n+2)} = \frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}$ for all $n \geq 1$.