Question

Aicharya · Accepted Answer

Consider $1^2+3^2+5^2+\ldots \ldots+25^2$ $n^{	ext {th }}$ 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}ight]+13-4 \frac{n(n+1)}{2} $ Put $n=13$, we get $ \begin{aligned} &\mathrm{S}_n=26 	imes 14 	imes 9+13-26 	imes 14 \ &=3276+13-364 \ &=2925 . \end{aligned} $

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