A finite series-2

Geometry Level 4

tan ⁡ [ ∑ n = 1 2021 tan ⁡ − 1 ( 1 x 2 + x + 1 ) ] = m n \tan\left [ \sum_{n=1}^{2021} \tan^{-1} \left (\frac{1}{x^2+x+1} \right ) \right ]=\frac{m}{n}

m m and n n are coprime positive integers satisfying the equation above. Find m − n . m-n.


The answer is -2.

Rafsan Rcc by Rafsan Rcc , 17, Bangladesh

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2 solutions

Rafsan Rcc
May 11, 2021

0 Helpful 0 Interesting 1 Brilliant 0 Confused
Dwaipayan Shikari
May 12, 2021

∑ x = 1 n tan ⁡ − 1 ( 1 x 2 + x + 1 ) = ∑ x = 1 n tan ⁡ − 1 ( x + 1 − x 1 + x ( 1 + x ) ) = ∑ x = 1 n ( tan ⁡ − 1 ( x + 1 ) − tan ⁡ − 1 ( x ) ) \displaystyle\small\sum_{x=1}^n \tan^{-1}\left(\dfrac{1}{x^2+x+1}\right) = \sum_{x=1}^n \tan^{-1}\left(\dfrac{x+1-x}{1+x(1+x)} \right)= \sum_{x=1}^n (\tan^{-1}(x+1)-\tan^{-1}(x))

∑ x = 1 n ( tan ⁡ − 1 ( x + 1 ) − tan ⁡ − 1 ( x ) ) = tan ⁡ − 1 ( 2 ) − tan ⁡ − 1 ( 1 ) + tan ⁡ − 1 ( 3 ) − ⋯ + tan ⁡ − 1 ( n + 1 ) − tan ⁡ − 1 ( n ) \displaystyle\small\sum_{x=1}^n (\tan^{-1}(x+1)-\tan^{-1}(x)) = \tan^{-1}(2)-\tan^{-1}(1)+\tan^{-1}(3)-\cdots+ \tan^{-1}(n+1)-\tan ^{-1}(n) = tan ⁡ − 1 ( n + 1 ) − tan ⁡ − 1 ( 1 ) = tan ⁡ − 1 ( n n + 2 ) \displaystyle=\tan^{-1}(n+1) -\tan^{-1}(1) = \tan^{-1}\left(\dfrac{n}{n+2}\right)

Here n = 2021 n=2021 , so therefore sum is tan ⁡ − 1 ( 2021 2023 ) \displaystyle \tan^{-1}(\dfrac{2021}{2023})

So tan ⁡ ( tan ⁡ − 1 ( 2021 2023 ) ) = 2021 2023 \tan\left(\tan^{-1}(\dfrac{2021}{2023})\right) = \dfrac{2021}{2023}

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