The sum of the following series $1+6+\frac{9\left(1^2+2^2+3^2\right)}{7}+\frac{12\left(1^2+2^2+3^2+4^2\right)}{9}$$+\frac{15\left(1^2+2^2+\dots +5^2\right)}{11}+\dots$ upto 15 terms, is

Solution:

The $n^{\text{th }}$ term of given expression is

$T_n=\frac{(3+(n-1\left)3\right)\left(1^2+2^2+3^2+\dots +n^2\right)}{(2n+1)}$

$=\frac{3n\cdot \frac{n(n+1)(2n+1)}{6}}{2n+1}=\frac{n^2(n+1)}{2}$

Now, $S_{15}=\frac{1}{2}\sum _{n=1}^{15} \left(n^3+n^2\right)$ $=\frac{1}{2}\left[{\left(\frac{15(15+1)}{2}\right)}^2+\frac{15\times 16\times 31}{6}\right]$

$=\frac{1}{2}\left[(120{)}^2+1240\right]=7820$