For any real number x, let cot -1 x denote the unique real number θ in (0, π) such that cot θ=x If undersetn arrow ∞ textLim displaystyle∑k=1n cot -1(1+k+k2)= cot -1(α)+ cot -1(β), where α, β are prime numbers, then find (α+β).

Question

For any real number x, let cot -1 x denote the unique real number θ in (0, π) such that cot θ=x If undersetn arrow ∞ textLim displaystyle∑k=1n cot -1(1+k+k2)= cot -1(α)+ cot -1(β), where α, β are prime numbers, then find (α+β). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

displaystyle∑k=1n cot -1(1+k+k2)= tan -1(n+1)- tan -1 1 take undersetn arrow ∞ textLim , we get R.H.S. =(π/2)-(π/4)=(π/4)= tan -1((1/α))+ tan -1((1/β)) ⇒(α-1)(β-1)=2 ∴ α=3, β=2 text (or α=2, β=3) ∴ text Answer =5

Q. For any real number x, let cot−1x denote the unique real number θ in (0,π) such that cotθ=x If n→∞Lim​k=1∑n​cot−1(1+k+k2)=cot−1(α)+cot−1(β), where α,β are prime numbers, then find (α+β).

k=1∑n​cot−1(1+k+k2)=tan−1(n+1)−tan−11 take n→∞Lim​, we get R.H.S. =2π​−4π​=4π​=tan−1(α1​)+tan−1(β1​) ⇒(α−1)(β−1)=2 ∴α=3,β=2 (or α=2,β=3) ∴ Answer =5

Q. For any real number $x$ , let $cot^{- 1} x$ denote the unique real number $θ$ in $(0, π)$ such that $cot θ = x$
If $n \to \infty Lim k = 1 \sum n cot^{- 1} (1 + k + k^{2}) = cot^{- 1} (α) + cot^{- 1} (β)$ , where $α, β$ are prime numbers, then find $(α + β)$ .