Question

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

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

Answer 1

The correct option is C $188090$

We have,

$t_n=[nth$ term of $1,2,3,...]\times [nth$ term of $3,5,7,...{]}^2$

$=n{(2n+1)}^2=4n^3+4n^2+n$

$\therefore S_n=\sum t_n=4\sum n^3+4\sum n^2+\sum n$

$=4.{\left\{\frac{n(n+1)}{2}\right\}}^2+4.\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$

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

$\therefore S_{20}={20}^2.{21}^2+\frac{2}{3}.20.21.41+\frac{1}{2}.20.21=188090$