Question

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

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

Answer 1

The correct option is B $7820$

$S=1+6+\frac{9(1^2+2^2+3^2)}{7}+\frac{12(1^2+2^2+3^2+4^2)}{9}+\frac{15(1^2+2^2+...+5^2)}{11}+...\text{up to}15\text{terms}$

$\Rightarrow S=\frac{3}{3}\left(1^2\right)+\frac{6}{5}(1^2+2^2)+\frac{12}{9}(1^2+2^2+3^2+4^2)+...+\text{up to}15\text{terms}$

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

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

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

$\Rightarrow S_n=\frac{1}{2}[\sum n^2+\sum n^3]$

$\Rightarrow S_n=\frac{1}{2}[\frac{n(n+1)(2n+1)}{6}+\frac{n^2(n+1{)}^2}{4}]$

$\Rightarrow S_n=\frac{1}{2}n(n+1)[\frac{2n+1}{6}+\frac{n(n+1)}{4}]$

$\therefore S_{15}=\frac{1}{2}\times 15(15+1)[\frac{31}{6}+\frac{15\times 16}{4}]$

$=\frac{1}{2}\times 15\times 16\left[\frac{62+720}{12}\right]$

$\Rightarrow S_{15}=7820$