Question

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

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

Answer 1

Answer: (A)

$T_n=\frac{\left(3+\left(n-1\right)\times 3\right)\left(1^2+2^2+....+n^2\right)}{\left(2n+1\right)}$
$T_n=\frac{3.\frac{n\left(n+1\right)\left(2n+1\right)}{6}}{2n+1}=\frac{n^2\left(n+1\right)}{2}$
$S_{15}=\frac{1}{2}{\sum }_{n=1}^{15}\left(n^3+n^2\right)=\frac{1}{2}\left[{\left(\frac{15\left(15+1\right)}{2}\right)}^2+\frac{15\times 16\times 31}{6}\right]$
$=7820$