Mathematics · 2024
JEE Main · 4 April 2024, Shift 2 · Q5
The value of (1 × 2^2+2 × 3^2+….+100 ×(101)^2)/(1^2 × 2+2^2 × 3+….+100^2 × 101) is
The value of $\displaystyle \frac{1 \times 2^2+2 \times 3^2+\ldots .+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots .+100^2 \times 101}$ is
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle \frac{305}{301}$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.