JEE PYQ: Sequence And Series Question 52
Question 52 - 2019 (10 Apr Shift 1)
The sum $\frac{3 \times 1^3}{1^2} + \frac{5 \times (1^3 + 2^3)}{1^2 + 2^2} + \frac{7 \times (1^3 + 2^3 + 3^3)}{1^2 + 2^2 + 3^2} + \ldots$ upto $10^{th}$ term, is:
(1) 680
(2) 600
(3) 660
(4) 620
Show Answer
Answer: (3)
Solution
$r^{th}$ term of the series, $T_r = \frac{(2r+1)\left(\frac{r(r+1)}{2}\right)^2}{\frac{r(r+1)(2r+1)}{6}} = \frac{3r(r+1)}{2}$
$\therefore$ sum of 10 terms is $= S = \sum T_r = \frac{3}{2}\sum(r^2 + r)$
$= \frac{3}{2}\left[\frac{10(10+1)(2 \times 10+1)}{6} + \frac{10(10+1)}{2}\right]$
$= \frac{3}{2} \times 5 \times 11 \times 8 = 660$
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