JEE PYQ: Sequence And Series Question 53
Question 53 - 2019 (10 Apr Shift 2)
The sum $1 + \frac{1^3 + 2^3}{1 + 2} + \frac{1^3 + 2^3 + 3^3}{1 + 2 + 3} + \ldots + \frac{1^3 + 2^3 + 3^3 + \ldots + 15^3}{1 + 2 + 3 + \ldots + 15} - \frac{1}{2}(1 + 2 + 3 + \ldots + 15)$ is equal to:
(1) 620
(2) 1240
(3) 1860
(4) 660
Show Answer
Answer: (1)
Solution
Let $S = \frac{1}{2}\sum_{n=1}^{15}\frac{m(n+1)}{2} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{2}$
$S = \frac{1}{2}\left(\sum_{n=1}^{15}n^2 + \sum_{n=1}^{15}n\right) = \frac{1}{2}\left(\frac{15(16)(31)}{6} + \frac{15(16)}{2}\right)$
$= 680$
$\therefore$ required sum is, $680 - \frac{1}{2} \cdot \frac{15(16)}{2} = 680 - 60 = 620$
BETA
