Answer
The required solution is ${{S}_{k}}=2{{k}^{2}}+2,{{S}_{k+1}}=\left( 2{{k}^{2}}+6k+4 \right)$
Work Step by Step
We are using ${{S}_{k}},{{S}_{k+1}}$ from the expression ${{S}_{n}}:4+8+12+...+4n=2n\left( n+1 \right)$
${{S}_{k}}$ is given by,
$\begin{align}
& 4+8+12+...+4k=2k\left( k+1 \right) \\
& =2{{k}^{2}}+2
\end{align}$
${{S}_{k+1}}$ Is provided by:
$\begin{align}
& 4+8+12+...+4\left( k+1 \right)=2\left( k+1 \right)\left( k+1+1 \right) \\
& =2\left( k+1 \right)\left( k+2 \right) \\
& =2\left( {{k}^{2}}+3k+2 \right) \\
& =\left( 2{{k}^{2}}+6k+4 \right)
\end{align}$
