Question

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

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

Answer 1

Answer: (C)

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