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

Question

If displaystyle∑n=1∞ tan -1((2n/22 n+1+3 ⋅ 2n+2))= cot -1(λ) then λ is equal to (A) 2 (B) 3 (C) 5 (D) 6

Tardigrade Admin · Accepted Answer

displaystyle∑n=1∞ tan -1((2n/22 n+1+3 ⋅ 2n+2))= displaystyle∑n=1∞ tan -1((2n/1+2 ⋅ 22 n+3 ⋅ 2n+1)) = displaystyle∑n=1∞ tan -1(((2n+1+1)-(2n+1)/1+(2n+1+1)(2n+1)))= displaystyle∑n=1∞( tan -1(2n+1+1)- tan -1(2n+1)) =(π/2)- tan -1 3= cot -1(3) ≡ cot -1(λ) ∴ λ=3 .

Q. If n=1∑∞​tan−1(22n+1+3⋅2n+22n​)=cot−1(λ) then λ is equal to

2

3

5

6

n=1∑∞​tan−1(22n+1+3⋅2n+22n​)=n=1∑∞​tan−1(1+2⋅22n+3⋅2n+12n​) =n=1∑∞​tan−1(1+(2n+1+1)(2n+1)(2n+1+1)−(2n+1)​)=n=1∑∞​(tan−1(2n+1+1)−tan−1(2n+1)) =2π​−tan−13=cot−1(3)≡cot−1(λ) ∴λ=3.

Q. If $n = 1 \sum \infty tan^{- 1} (\frac{2 ^{n}}{2 ^{2 n + 1} + 3 \cdot 2 ^{n} + 2}) = cot^{- 1} (λ)$ then $λ$ is equal to