Question

Let S_k = `sum_(r = 1)^k tan^-1(6^r/(2^(2r + 1) + 3^(2r + 1)))`. Then `lim_(k→∞)` S_k = is equal to ______.

Options

• `tan^-1(3/2)`

• `cot^-1(3/2)`

• `π/2`

• `tan^-1(3)`

Answer 1

Let S_k = `sum_(r = 1)^k tan^-1(6^r/(2^(2r + 1) + 3^(2r + 1)))`. Then `lim_(k→∞)` S_k = is equal to `underlinebb(cot^-1(3/2))`.

Explanation:

Given, S_k = `sum_(r = 1)^k tan^-1(6^r/(2^(2r + 1) + 3^(2r + 1)))`

⇒ S_k = `sum_(r = 1)^k tan^-1((6^r/9^r)/((2^(2r + 1) + 3^(2r + 1))/9^r))`

⇒ S_k = `sum_(r = 1)^k tan^-1{(((2^r.3^r)/(3^r.3^r)))/((2/3)^(2r).2+ ((3^(2r))/(3^(2r))).3)}`

⇒ S_k = `sum_(r = 1)^k tan^-1{(2/3)^r/(3 + 2.(2/3)^(2r))}`

⇒ S_k = `sum_(r = 1)^k tan^-1{(2/3)^r/(3(1 + (2/3).(2/3)^(2r)))}`

Let `(2/3)^r` = t

⇒ S_k = `sum_(r = 1)^k tan^-1{t/(3.(1 + (2/3)t^2))}`

⇒ S_k = `sum_(r = 1)^k tan^-1{(t /3)/(1 + (t).((2t)/3))}`

⇒ S_k = `sum_(r = 1)^k tan^-1{(t - (2t)/3)/(1 + (t).((2t)/3))}`

⇒ S_k = `sum_(r = 1)^k {tan^-1(t) - tan^-1((2t)/3)}`

⇒ S_k = `sum_(r = 1)^k {tan^-1(2/3)^r - tan^-1(2/3)^(r + 1)}`

⇒ S_k = `{tan^-1(2/3) - tan^-1(2/3)^2} + {tan^-1(2/3)^2 - tan^-1(2/3)^3} + {tan^-1(2/3)^3 - tan^-1(2/3)^4} + -------- + {tan^-1(2/3)^k - tan^-1(2/3)^(k + 1)}`

⇒ S_k = `tan^-1(2/3) - tan^-1(2/3)^(k + 1)`

⇒ `lim_(k→∞)(S_k) = lim_(k→∞){tan^-1(2/3) - tan^-1(2/3)^(k + 1)}`

⇒ S_∞ = `tan^-1(2/3) - tan^-1(0); lim_(k→∞)(2/3)^(k + 1)` = 0

⇒ S_∞ = `tan^-1(2/3) - 0`

⇒ S_∞ = `tan^-1(2/3)`

⇒ S_∞ = `cot^-1(3/2)`