Question

`lim_(n→∞)tan{sum_(r = 1)^n tan^-1(1/(1 + r + r^2))}` is equal to ______. 

पर्याय

• 0

• 1

• 2

• 3

Answer 1

`lim_(n→∞)tan{sum_(r = 1)^n tan^-1(1/(1 + r + r^2))}` is equal to 1. 

Explanation:

Given: `lim_(n→∞)tan{sum_(r = 1)^n tan^-1(1/(1 + r + r^2))}`

Let T_r = `tan^-1(1/(1 + r + r^2))`

= `tan^-1(1/(1 + r(r + 1)))`

1 can be written as (r + 1) – r

T_r = `tan^-1(((r + 1) - r)/(1 + (r + 1)r))`  ...`(∵ tan^-1((x - y)/(1 + xy)) = tan^-1x - tan^-1y)`

Tr = tan^–1(r + 1) – tan^–1(r)

T₁ = tan^–1(2) – tan^–1(1)

T₂ = tan^–1(3) – tan^–1(2)

T₃ = tan^–1(4) – tan^–1(3)
.
.
.
.
.
.
T_n = tan^–1(n + 1) – tan^–1(n)

Adding all, we get

`sum_(r = 1)^n tan^-1(1/(1 + r + r^2))` = tan^–1(n + 1) – tan^–1(1)  ...(i)

= `tan^-1(((n + 1) - 1)/(1 + n + 1))`

= `tan^-1(n/(n + 2))`

= `tan^-1(1/(1 + 2/n))`

`lim_(n→∞)sum_(r = 1)^n tan^-1(1/(1 + r + r^2)) = tan^-1(1) = π/4`

`lim_(n→∞)tan{sum_(r = 1)^n tan^-1(1/(1 + r + r^2))} = tan  π/4` = 1