[JEE Advanced 1993] Prove by induction that $\tan^{-1}\frac13+\tan^{-1}\frac17+\cdots+\tan^{-1}\frac1{n^2+n+1}=\tan^{-1}\frac n{n+2}$.
Prove by induction that $\tan^{-1}\frac13+\tan^{-1}\frac17+\cdots+\tan^{-1}\frac1{n^2+n+1}=\tan^{-1}\frac n{n+2}$.
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VAVidaara Admin
✓ Vidaara Team
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· 1mo ago
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Answer: Proved.
Each term $\tan^{-1}\frac1{n^2+n+1}=\tan^{-1}(n+1)-\tan^{-1}n$ telescopes; with $\tan^{-1}(n+1)-\tan^{-1}1=\tan^{-1}\frac n{n+2}$.
JEE Advanced 1993 · Binomial Theorem — verified solution by the Vidaara Team.
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