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Mathematics · 2021

JEE Main · 26 February 2021, Shift 2 · Q64

If 0<a, b<1, and tan^-1 a+ tan^-1 b=(π)/4, then the value of ( a + b )-((a^2+ b^2)/2)+((a^3+ b^3)/3)-((a^4+ b^4)/4)+… is:

If $\displaystyle 0<a, b<1$, and $\displaystyle \tan ^{-1} a+\tan ^{-1} b=\frac{\pi}{4}$, then the value of $\displaystyle (\mathrm{a}+\mathrm{b})-\left(\frac{\mathrm{a}^2+\mathrm{b}^2}{2}\right)+\left(\frac{\mathrm{a}^3+\mathrm{b}^3}{3}\right)-\left(\frac{\mathrm{a}^4+\mathrm{b}^4}{4}\right)+\ldots$ is :
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.