Question

The value of  `lim_(n→∞)1/n sum_(r = 0)^(2n-1) n^2/(n^2 + 4r^2)` is ______.

विकल्प

• `1/4 tan^-1(4)`

• tan^–1(4)

• `1/4 tan^-1(2)`

• `1/2 tan^-1(4)`

Answer 1

The value of  `lim_(n→∞)1/n sum_(r = 0)^(2n-1) n^2/(n^2 + 4r^2)` is `underlinebb(1/2 tan^-1(4))`.

Explanation:

⇒ I = `lim_(n→∞)1/n sum_(r = 0)^(2n-1) n^2/(n^2 + 4r^2)`

⇒ I = `lim_(n→∞)1/n sum_(r = 0)^(2n-1) 1/n 1/(1 + 4(r/n)^2`

Put, `r/n→x, 1/n→dx, lim_(n→∞) 0/n` = 0, `lim_(n→∞) (2n - 1)/n` = 2

I = `int_0^2 1/(1 + 4x^2)dx`

⇒ I = `1/4int_0^2 (dx)/(1/4 + x^2)`

= `1/4 .2[tan^-1 2x]_0^2`

= `1/2tan^-1(4)`