Answer
The required solution is ${{S}_{k}}=k\left( 2k+1 \right),{{S}_{k+1}}=(k+1)(2k+3)$
Work Step by Step
Then using ${{S}_{k}},{{S}_{k+1}}$ from the expression ${{S}_{n}}:3+7+11+...+\left( 4n-1 \right)=n\left( 2n+1 \right)$
${{S}_{k}}$ is provided by,
$3+7+11+...+\left( 4k-1 \right)=k\left( 2k+1 \right)$
${{S}_{k+1}}$ is given by,
$\begin{align}
& 3+7+11+...+\left( 4\left( k+1 \right)-1 \right)=\left( k+1 \right)\left( 2\left( k+1 \right)+1 \right) \\
& =\left( k+1 \right)\left( 2k+2+1 \right) \\
& =\left( k+1 \right)\left( 2k+3 \right) \\
\end{align}$
We get the solution ${{S}_{k}}=k\left( 2k+1 \right),{{S}_{k+1}}=(k+1)(2k+3)$.
