Sequences And Series Ques 32
- The sum of the following series $1+6+\frac{9\left(1^2+2^2+3^2\right)}{7}+\frac{12\left(1^2+2^2+3^2+4^2\right)}{9}$ $+\frac{15\left(1^2+2^2+\ldots+5^2\right)}{11}+\ldots$ up to $15$ terms is
(2019 Main, 9 Jan II)
(a) $7510$
(b) $7820$
(c) $7830$
(d) $7520$
Show Answer
Answer:
Correct Answer: 32.(b)
Solution: (b) General term of the given series is
$T_r=\frac{3 r\left(1^2+2^2+\ldots+r^2\right)}{2 r+1}=\frac{3 r[r(r+1)(2 r+1)]}{6(2 r+1)}$
$=\frac{1}{2}\left(r^3+r^2\right)$
Now, required sum $=\sum_{r=1}^{15} T_r=\frac{1}{2} \sum_{r=1}^{15}\left(r^3+r^2\right)$
$=\frac{1}{2}\left\{\left[\frac{n(n+1)}{2}\right]^2+\frac{n(n+1)(2 n+1)}{6}\right\}_{n-15}$
$=\frac{1}{2}\left\{\frac{n(n+1)}{2}\left[\frac{n^2+n}{2}+\frac{2 n+1}{3}\right]\right\}_{n-15}$
$=\frac{1}{2}\left\{\frac{n(n+1)}{2} \frac{\left(3 n^2+7 n+2\right)}{6}\right\}_{n-15}$
$=\frac{1}{2} \times \frac{15 \times 16}{2} \times \frac{(3 \times 225+105+2)}{6}=7820$
BETA
