Consider $1^2+3^2+5^2+\ldots \ldots+25^2$ $n^{\text {th }}$ term $\mathrm{T}_n=(2 n-1)^2, n=1, \ldots \ldots 13$
Now, $\mathrm{S}_n=\sum_{n=1}^{13} \mathrm{~T}_n=\sum_{n=1}^{13}(2 n-1)^2$
$$
\begin{aligned}
&=\sum_{n=1}^{13} 4 n^2+\sum_{n=1}^{13} 1-\sum_{n=1}^{13} 4 n \\
&=4 \sum n^2+13-4 \sum n
\end{aligned}
$$
$$
=4\left[\frac{n(n+1)(2 n+1)}{6}\right]+13-4 \frac{n(n+1)}{2}
$$
Put $n=13$, we get
$$
\begin{aligned}
&\mathrm{S}_n=26 \times 14 \times 9+13-26 \times 14 \\
&=3276+13-364 \\
&=2925 .
\end{aligned}
$$