Answer
The required solution is ${{S}_{k}}=\frac{k\left( 5k-1 \right)}{2},{{S}_{k+1}}=\frac{\left( 5{{k}^{2}}+9k+4 \right)}{2}$
Work Step by Step
Now, using ${{S}_{k}},{{S}_{k+1}}$ from the expression ${{S}_{n}}:2+7+12+...+\left( 5n-3 \right)=\frac{n\left( 5n-1 \right)}{2}$
${{S}_{k}}$ is provided by,
$2+7+12+...+\left( 5k-3 \right)=\frac{k\left( 5k-1 \right)}{2}$
${{S}_{k+1}}$ is provided by,
$\begin{align}
& 2+7+12+...+\left( 5\left( k+1 \right)-3 \right)=\frac{\left( k+1 \right)\left( 5\left( k+1 \right)-1 \right)}{2} \\
& =\frac{\left( k+1 \right)\left( 5k+5-1 \right)}{2} \\
& =\frac{\left( k+1 \right)\left( 5k+4 \right)}{2} \\
& =\frac{\left( 5{{k}^{2}}+9k+4 \right)}{2}
\end{align}$
Then, we get the solution
${{S}_{k}}=\frac{k\left( 5k-1 \right)}{2},{{S}_{k+1}}=\frac{\left( 5{{k}^{2}}+9k+4 \right)}{2}$.
