Answer
.
Work Step by Step
We use mathematical induction as follows:
Statement ${{S}_{1}}$ is
${{S}_{1}}:4$
We are solving the right side as,
$\begin{align}
& 2n\left( n+1 \right)=2\left( 1 \right)\left( 1+1 \right) \\
& =2\left( 2 \right) \\
& =4
\end{align}$
This statement shows that ${{S}_{1}}$ is true.
Suppose ${{S}_{k}}$ is true. Using ${{S}_{k}},{{S}_{k+1}}$ from the expression,
$\begin{align}
& {{S}_{k}}:4+8+12+...+4k=2k\left( k+1 \right) \\
& \text{ }=2{{k}^{2}}+2 \\
\end{align}$
Then we have ${{S}_{k+1}}$
${{S}_{k+1}}:$
$\begin{align}
& 4+8+12+...+4\left( k+1 \right)=2\left( k+1 \right)\left( k+1+1 \right) \\
& =2\left( k+1 \right)\left( k+2 \right) \\
& =2\left( {{k}^{2}}+k+2k+2 \right) \\
& =\left( 2{{k}^{2}}+6k+4 \right)
\end{align}$
Solving further
$\begin{align}
& 4+8+12+...+4\left( k+1 \right)=2\left( {{k}^{2}}+3k+1 \right)+2 \\
& =2{{\left( k+1 \right)}^{2}}+2
\end{align}$
Thus, ${{S}_{k+1}}$ is true. The result ${{S}_{n}}:4+8+12+...+4n=2n\left( n+1 \right)$ is true by mathematical induction.
Hence, the provided statement is proved by mathematical induction.
