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 $$\begin{align}&\left(\frac1{1\cdot 3}+\frac1{2\cdot 4}+\cdots+\frac1{n(n+2)}\right)+\frac1{(n+1)(n+3)}\\ &=\frac34-\frac{2n+3}{2(n+1)(n+2)}+\frac1{(n+1)(n+3)} \end{align}$$ and you should obtain $(*)$ with straightforward computings.
Perhaps my explanation is not very brief but I hope it to be useful.
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*}
Hint: $$\frac { 1 }{ 2 } \left( \frac { 1 }{ n } -\frac { 1 }{ n+2 } \right) =\frac { 1 }{ n\left( n+2 \right) } $$
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 $p(k+1).$
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.
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 $n\geq 1$. $\blacksquare$
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 $1$.
$$\begin{align} \mathrm{LHS} &= \frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+\cdots+\frac{1}{(n+1)(n+3)} \\ &= \frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+\cdots+\frac{1}{n(n+2)}+\frac{1}{(n+1)(n+3)} \\ &= \left( \frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+\cdots+\frac{1}{n(n+2)}\right)+\frac{1}{(n+1)(n+3)} \\ &= \frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}+\frac{1}{(n+1)(n+3)} \\ &= \frac{3}{4}-\frac{(2n+3)(n+3)}{2(n+1)(n+2)(n+3)}+\frac{2(n+2)}{2(n+1)(n+2)(n+3)} \\ &= \frac{3}{4}+\frac{-(2n+3)(n+3)+2(n+2)}{2(n+1)(n+2)(n+3)} \\ &= \frac{3}{4}+\frac{-(2n^2+6n+3n+9)+2n+4}{2(n+1)(n+2)(n+3)} \\ &= \frac{3}{4}+\frac{-2n^2-9n-9+2n+4}{2(n+1)(n+2)(n+3)} \\ &= \frac{3}{4}+\frac{-2n^2-7n-5}{2(n+1)(n+2)(n+3)} \\ &= \frac{3}{4}-\frac{2n^2+7n+5}{2(n+1)(n+2)(n+3)} \\ &= \frac{3}{4}-\frac{2n^2+2n+5n+5}{2(n+1)(n+2)(n+3)} \\ &= \frac{3}{4}-\frac{2n(n+1)+5(n+1)}{2(n+1)(n+2)(n+3)} \\ &= \frac{3}{4}-\frac{(n+1)(2n+5)}{2(n+1)(n+2)(n+3)} \\ &= \frac{3}{4}-\frac{2n+5}{2(n+2)(n+3)} \\ &= \frac{3}{4}-\frac{2n+2+3}{2(n+1+1)(n+1+2)} \\ &= \frac{3}{4}-\frac{2(n+1)+3}{2((n+1)+1)((n+1)+2)}. \end{align}$$
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 $n \geq 1$.
