Mathematics · 2021
JEE Main · 27 August 2021, Shift 1 · Q78
If 0<x<1, then 3/2 x^2+5/3 x^3+7/4 x^4+…, is equal to:
If $\displaystyle 0<x<1$, then $\displaystyle \frac{3}{2} x^2+\frac{5}{3} x^3+\frac{7}{4} x^4+\ldots$, is equal to:
Official answer
(1)
$\displaystyle x\left(\frac{1+x}{1-x}\right)+\log _{\mathrm{e}}(1-x)$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.