Question

Sum to $20$ terms of the series $1.3^2+2.5^2+3.7^2+\dots$ is

• A. 178090
• B. 168090
• C. 188090
• D. None of these

Answer 1

Answer: (C)

We have,
$t_n=[n\text{th term of }1,2,3,\dots ]\times [n\text{th term of }3,5,7,\dots {]}^2$
$=n(2n+1{)}^2=4n^3+4n^2+n$
$\therefore S_n=\Sigma t_n=4\Sigma n^3+4\Sigma n^2+\Sigma n$
$=4\cdot {\left[\frac{n(n+1)}{2}\right]}^2+4\cdot \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$
$=n^2(n+1{)}^2+\frac{2}{3}n(n+1)(2n+1)+\frac{1}{2}n(n+1)$
$\therefore S_{20}=20^2\cdot 21^2+\frac{2}{3}\times 20\cdot 21\cdot 41+\frac{1}{2}\cdot 20\cdot 21$
$=188090$