Answer
See the explanation below.
Work Step by Step
We use mathematical induction as follows:
Statement ${{S}_{1}}$ is:
${{S}_{1}}:3$
Then, simplify on the right and obtain,
$\begin{align}
& n\left( 2n+1 \right)=1\left( 2+1 \right) \\
& =3
\end{align}$.
This statement shows that ${{S}_{1}}$ is true.
Suppose ${{S}_{k}}$ is true. By using ${{S}_{k}},{{S}_{k+1}}$ from the expression,
${{S}_{k}}:3+7+11+...+\left( 4k-1 \right)=k\left( 2k+1 \right)$
Add $\left( 4\left( k+1 \right)-1 \right)$ on both sides,
$\begin{align}
& 3+7+11+...+\left( 4k-1 \right)+\left( 4\left( k+1 \right)-1 \right)=k\left( 2k+1 \right)+\left( 4\left( k+1 \right)-1 \right) \\
& =2{{k}^{2}}+5k+3 \\
& =2{{k}^{2}}+3k+2k+3 \\
& =k\left( 2k+3 \right)+1\left( 2k+3 \right)
\end{align}$
Simplify
$\begin{align}
& 3+7+11+...+\left( 4k-1 \right)+\left( 4\left( k+1 \right)-1 \right)=\left( k+1 \right)\left( 2k+3 \right) \\
& =\left( k+1 \right)\left( 2\left( k+1 \right)+1 \right)
\end{align}$
Therefore, ${{S}_{k+1}}$ is true. The result ${{S}_{n}}:3+7+11+...+\left( 4n-1 \right)=n\left( 2n+1 \right)$ is true by mathematical induction.
Thus, the provided statement is proved by mathematical induction.
