354. Find the expressions for the sums:

\[
\begin{aligned}
& 2^{2}+4^{2}+\ldots+(2 n)^{2} \\
& 1+3^{2}+\ldots+(2 n+1)^{2}
\end{aligned}
\]
Solution. Let's transform the expression:

$$
\begin{gathered}
2^{2}+4^{2}+\ldots+4 n^{2}=4\left(1+2^{2}+\ldots+n^{2}\right)= \\
=\frac{2 n(n+1)(2 n+1)}{3}
\end{gathered}
$$

To determine the sum $1+3^{2}+\ldots+(2 n+1)^{2}$, consider the sum:

$$
\begin{gathered}
1+2^{2}+3^{2}+\ldots+n^{2}+\ldots+(2 n+1)^{2}= \\
=\frac{(2 n+1)(2 n+2)(4 n+3)}{6}
\end{gathered}
$$

(according to problem 353). Therefore,

$$
\begin{gathered}
1+3^{2}+\ldots+(2 n+1)^{2}=\frac{(2 n+1)(2 n+2)(4 n+3)}{6}-\frac{2 n(n+1)(2 n+1)}{3}= \\
=\frac{(n+1)(2 n+1)(2 n+3)}{3}
\end{gathered}
$$