Answer
See the explanation below.
Work Step by Step
We use mathematical induction as follows:
Statement ${{S}_{1}}$ is
${{S}_{1}}:2$
And simplifying on the right, we obtain,
$\begin{align}
& \frac{n\left( 5n-1 \right)}{2}=\frac{1\left( 5-1 \right)}{2} \\
& =\frac{4}{2} \\
& =2
\end{align}$.
So, this statement shows that ${{S}_{1}}$ is true.
And assume ${{S}_{k}}$ is true. Using ${{S}_{k}},{{S}_{k+1}}$ from the expression,
${{S}_{k}}:2+7+12+...+\left( 5k-3 \right)=\frac{k\left( 5k-1 \right)}{2}$
Add $\left( 5\left( k+1 \right)-3 \right)$ on both sides,
$\begin{align}
& 2+7+12+...+\left( 5k-3 \right)+\left( 5\left( k+1 \right)-3 \right)=\frac{k\left( 5k-1 \right)}{2}+\left( 5\left( k+1 \right)-3 \right) \\
& =\frac{k\left( 5k-1 \right)}{2}+\left( 5k+2 \right) \\
& =\frac{5{{k}^{2}}+9k+4}{2} \\
& =\frac{\left( 5{{k}^{2}}+4k+5k+4 \right)}{2}
\end{align}$
We simplify further, $\begin{align}
& 2+7+12+...+\left( 5k-3 \right)+\left( 5\left( k+1 \right)-3 \right)=\frac{k\left( 5k+4 \right)+1\left( 5k+4 \right)}{2} \\
& =\frac{\left( k+1 \right)\left( 5k+4 \right)}{2} \\
& =\frac{\left( k+1 \right)\left( 5\left( k+1 \right)-1 \right)}{2}
\end{align}$
Therefore, ${{S}_{k+1}}$ is true. The result ${{S}_{n}}:2+7+12+...+\left( 5n-3 \right)=\frac{n\left( 5n-1 \right)}{2}$ is true by mathematical induction.
Thus, the provided statement is proved by mathematical induction.
