Question

The sum of the following series $1+6+\frac{9(1^2+2^2+3^2)}{7}+\frac{12(1^2+2^2+3^2+4^2)}{9}+\frac{15(1^2+2^2+....+5^2)}{11}+.....$ up to $15$ terms is:

• A. $7820$
• B. $7830$
• C. $7520$
• D. $7510$

Answer 1

The correct option is A $7820$

$T_n=\frac{(3+(n-1)\times 3)(1^2+2^2+...+n^2)}{(2n+1)}$

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

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

$=7820$