If $\frac{\pi}{4}+\sum_{p=1}^{11} \tan ^{-1}\left(\frac{2^{p-1}}{1+2^{2 p-1}}\right)=\alpha$, then $\tan \alpha$ is equal to ______.

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Inverse Trigonometric Functions

\frac{π}{4} + \sum_{p = 1}^{11} tan^{- 1} (\frac{2 ^{p - 1}}{1 + 2 ^{2 p - 1}}) = α

, then

tan α

is equal to ______.

Answer: 2048

Writing

2^{p - 1} = 2^{p} - 2^{p - 1}

and

1 + 2^{2 p - 1} = 1 + 2^{p} \cdot 2^{p - 1}

, the general term becomes

tan^{- 1} (\frac{2 ^{p} - 2 ^{p - 1}}{1 + 2 ^{p} \cdot 2 ^{p - 1}}) = tan^{- 1} (2^{p}) - tan^{- 1} (2^{p - 1})

. The sum therefore telescopes to

tan^{- 1} (2^{11}) - tan^{- 1} (2^{0}) = tan^{- 1} (2048) - \frac{π}{4}

. Adding the leading

\frac{π}{4}

gives

α = tan^{- 1} (2048)

, so

tan α = 2048

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