Answer
The required solution is $\left( k+1 \right)\left( 2k+3 \right)$.
Work Step by Step
Let us consider the statement $3+7+11+\ldots +\left( 4n-1 \right)=n\left( 2n+1 \right)$
If $ n=1$, the statement is $3=1\left( 2+1 \right)$
If $ n=2$, the statement is $3+7=2\left( 4+1 \right)$
If $ n=3$, the statement is $3+7+11=3\left( 6+1 \right)$.
If $ n=k+1$, the statement before the algebra is simplified is $3+7+11+\ldots +\left( 4\left( k+1 \right)-1 \right)=\left( k+1 \right)\left( 2\left( k+1 \right)+1 \right)$.
If $ n=k+1$, the statement after the algebra is simplified is $3+7+11+\ldots +\left( 4k+3 \right)=\left( k+1 \right)\left( 2k+3 \right)$.
For the statement $3+7+11+\ldots +\left( 4n-1 \right)=n\left( 2n+1 \right)$, and if $ n=1$ putting,
$\begin{align}
& 3+7+11+\ldots +\left( 4n-1 \right)=n\left( 2n+1 \right) \\
& 3=1\left( 2\times 1+1 \right) \\
& 3=1\left( 2+1 \right)
\end{align}$
If the value of $ n=2$ then we get,
$\begin{align}
& 3+7+11+\ldots +\left( 4n-1 \right)=n\left( 2n+1 \right) \\
& 3+\ldots +\left( 4\times 2-1 \right)=2\left( 2\times 2+1 \right) \\
& 3+\ldots +\left( 8-1 \right)=2\left( 4+1 \right) \\
& 3+7=2\left( 4+1 \right)
\end{align}$
After that, putting $ n=3$, we get
$\begin{align}
& 3+7+11+\ldots +\left( 4n-1 \right)=n\left( 2n+1 \right) \\
& 3+\ldots +\left( 4\times 3-1 \right)=3\left( 2\times 3+1 \right) \\
& 3+\ldots +\left( 12-1 \right)=3\left( 6+1 \right) \\
& 3+7+11=3\left( 5+1 \right)
\end{align}$
If $ n=k+1$ substituting,
$3+7+11+\ldots +\left( 4k-1 \right)=k\left( 2k+1 \right)$
Further simplification gives,
$\begin{align}
& 3+7+11+\ldots +\left( 4\left( k+1 \right)-1 \right)=\left( k+1 \right)\left( 2\left( k+1 \right)+1 \right) \\
& 3+7+11+\ldots +\left( 4k+4-1 \right)=\left( k+1 \right)\left( 2k+2+1 \right) \\
& 3+7+11+\ldots +\left( 4k+3 \right)=\left( k+1 \right)\left( 2k+3 \right)
\end{align}$
