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  4. If displaystyle∑n=1∞ tan -1((2n/22 n+1+3 ⋅ 2n+2))= cot -1(λ) then λ is equal to

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Q. If $\displaystyle\sum_{n=1}^{\infty} \tan ^{-1}\left(\frac{2^n}{2^{2 n+1}+3 \cdot 2^n+2}\right)=\cot ^{-1}(\lambda)$ then $\lambda$ is equal to

Inverse Trigonometric Functions

Solution:

$\displaystyle\sum_{n=1}^{\infty} \tan ^{-1}\left(\frac{2^n}{2^{2 n+1}+3 \cdot 2^n+2}\right)=\displaystyle\sum_{n=1}^{\infty} \tan ^{-1}\left(\frac{2^n}{1+2 \cdot 2^{2 n}+3 \cdot 2^n+1}\right) $
$=\displaystyle\sum_{n=1}^{\infty} \tan ^{-1}\left(\frac{\left(2^{n+1}+1\right)-\left(2^n+1\right)}{1+\left(2^{n+1}+1\right)\left(2^n+1\right)}\right)=\displaystyle\sum_{n=1}^{\infty}\left(\tan ^{-1}\left(2^{n+1}+1\right)-\tan ^{-1}\left(2^n+1\right)\right) $
$=\frac{\pi}{2}-\tan ^{-1} 3=\cot ^{-1}(3) \equiv \cot ^{-1}(\lambda) $
$\therefore \lambda=3 .$