15. Given a positive integer $n$ and a positive number $M$. For all arithmetic sequences $a_{1}, a_{2}, a_{3}, \cdots$ satisfying the condition $a_{1}^{2}+a_{n+1}^{2} \leqslant M$, find the maximum value of $S=a_{n+1}+a_{n+2}+\cdots+a_{2 n+1}$.
$$
\begin{array}{l}
S=a_{n+1}+a_{n+2}+\cdots+a_{2 n+1}=\frac{(n+1)\left(a_{n+1}+a_{2 n+1}\right)}{2}=\frac{(n+1)\left(3 a_{n+1}-a_{1}\right)}{2} \\
\leqslant \frac{(n+1) \sqrt{\left[3^{2}+(-1)^{2}\right]\left(a_{n+1}^{2}+a_{1}^{2}\right)}}{2} \leqslant \frac{(n+1) \sqrt{10 M}}{2} . \\
\text { Equality holds when }\left\{\begin{array} { l } 
{ \frac { a _ { n + 1 } } { 3 } = \frac { a _ { 1 } } { - 1 } , } \\
{ 3 a _ { n + 1 } - a _ { 1 } = \sqrt { 1 0 M } }
\end{array} \Rightarrow \left\{\begin{array}{l}
a_{1}=-\frac{\sqrt{10 M}}{10}, \\
a_{n+1}=\frac{3 \sqrt{10 M}}{10} .
\end{array}\right.\right.
\end{array}
$$

Therefore, the maximum value of $S=a_{n+1}+a_{n+2}+\cdots+a_{2 n+1}$ is $\frac{(n+1) \sqrt{10 M}}{2}$.