Mathematics · 2019
JEE Main · 10 April 2019, Shift 1 · Q68
The sum (3 × 1^3)/1^2+(5 ×(1^3+2^3))/(1^2+2^2)+(7 ×(1^3+2^3+3^3))/(1^2+2^2+3^2)+… …. upto 10^th term, is:
The sum
$$\frac{3 \times 1^3}{1^2}+\frac{5 \times\left(1^3+2^3\right)}{1^2+2^2}+\frac{7 \times\left(1^3+2^3+3^3\right)}{1^2+2^2+3^2}+\ldots \ldots .
$$
upto $\displaystyle 10^{\text {th }}$ term, is :
Official answer
(3)
$\displaystyle 660$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.