Mathematics - Sequence & Series Question with Solution | TestHub

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