Mathematics · 2019
JEE Main · 10 April 2019, Shift 2 · Q69
The sum 1+(1^3+2^3)/(1+2)+(1^3+2^3+3^3)/(1+2+3)+…. +(1^3+2^3+3^3+…+15^3)/(1+2+3+…+15)-1/2(1+2+3+…+15) is equal to:
The sum $\displaystyle 1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+\ldots$.
$$+\frac{1^3+2^3+3^3+\ldots+15^3}{1+2+3+\ldots+15}-\frac{1}{2}(1+2+3+\ldots+15)
$$
is equal to :
Official answer
(1)
$\displaystyle 620$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.