Offline paper
Mathematics · 2016
JEE Main 2016 · 2016-04-03 Offline · Q69
If the sum of the first ten terms of the series (1 3/5)^2+(2 2/5)^2+(3 1/5)^2+4^2+(4 4/5)^2+… …, is 16/5 m, then m is equal to:
If the sum of the first ten terms of the series $\displaystyle \left(1 \frac{3}{5}\right)^2+\left(2 \frac{2}{5}\right)^2+\left(3 \frac{1}{5}\right)^2+4^2+\left(4 \frac{4}{5}\right)^2+\ldots \ldots$, is $\displaystyle \frac{16}{5} m$, then $\displaystyle m$ is equal to :
Official answer
(2)
$\displaystyle 101$
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JEE Main 2016 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.