#### Answer

See the full explanation below.

#### Work Step by Step

The expression is given below:
${{S}_{n}}=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\cdots +\frac{1}{2n\left( n+1 \right)}$
And the above sum can be written in a general from
${{S}_{n}}=\sum\limits_{n=1}^{n}{\frac{1}{2n\left( n+1 \right)}}$
It can be simplified as
$\begin{align}
& {{S}_{n}}=\frac{1}{2}\sum\limits_{n=1}^{n}{\frac{\left( 2n+2-2n \right)}{2n\left( n+1 \right)}} \\
& {{S}_{n}}=\frac{1}{2}\sum\limits_{n=1}^{n}{\frac{2\left( n+1 \right)-2n}{2n\left( n+1 \right)}} \\
\end{align}$
${{S}_{n}}=\frac{1}{2}\sum\limits_{n=1}^{n}{\left( \frac{1}{n}-\frac{1}{n+1} \right)}$
and by putting the values of n in the above equation
$\begin{align}
& {{S}_{n}}=\frac{1}{2}\left( \left( 1-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right)+\left( \frac{1}{3}-\frac{1}{4} \right)\ldots \left( \frac{1}{n}-\frac{1}{n+1} \right) \right) \\
& {{S}_{n}}=\frac{1}{2}\left( 1-\frac{1}{n+1} \right) \\
& {{S}_{n}}=\frac{n}{2\left( n+1 \right)} \\
\end{align}$ $(I)$
Now, prove the conjectural formula by mathematical induction
Let ${{S}_{n}}$ be a statement involving the positive integer n. If
1. ${{S}_{1}}$ is true, and
2. Let us assume that the ${{S}_{k}}$ be true for some integer $ k $.
Then, one has to use ${{S}_{k}}$ to prove ${{S}_{k+1}}$ is also true. If one is able to prove ${{S}_{k+1}}$ then by the principle of mathematical induction the statement ${{S}_{n}}$ is true for all positive integers $ n $.
For ${{S}_{1}}$ put $ n=1$ in the above equation, one gets
${{S}_{1}}=\frac{1}{4}$,
It is the same as the first term, thus the first equation is satisfied.
Then, let us assume that ${{S}_{k}}$ is true, then ${{S}_{k+1}}$ should also be true by the method of mathematical induction.
${{S}_{k}}=\frac{k}{2\left( k+1 \right)}$ $(II)$
And for $ n=k+1$
${{S}_{k+1}}=\frac{\left( k+1 \right)}{2\left( k+2 \right)}$
Now, one has
$\begin{align}
& {{S}_{k+1}}={{S}_{k}}+\left( \frac{1}{2(k+1)(k+2)} \right) \\
& {{S}_{k+1}}=\left( \frac{k}{2(k+1)} \right)+\left( \frac{1}{2(k+1)(k+2)} \right) \\
\end{align}$
And simplify
${{S}_{k+1}}=\frac{\left( k+1 \right)}{2\left( k+2 \right)}$
So, the inequality for $ k+1$ is the same as equation (I), which means it has satisfied the second condition of mathematical induction.
Thus, the expression ${{S}_{n}}=\frac{n}{2\left( n+1 \right)}$ is true for all positive integers $ n $.
And the value are ${{S}_{1}}=\frac{1}{4}$, ${{S}_{2}}=\frac{1}{3}$, ${{S}_{3}}=\frac{3}{8}$, and ${{S}_{5}}=\frac{5}{12}$