Question

can anyone send me solution of this​

Answer 1

Answer:

π/6

Step-by-step explanation:

use definite integral for a limit of a sum

S = 1/ √( 4n² - 1 ) + 1/ √( 4n² - 2) +.........

S = 1/ 2n √( 1 - 1/4n² ) + 1/2n √( 1 - 2/4n²) +.........

S = 1/2n [1/(1 - 1/4n² ) + 1/√( 1 - 2/4n²) +.........]

S = 1/2n [ 1/(1 - (1/2n)² +1/(1 - (2/2n)²+..]

S = 1/2× 1/n €r (1, ∞) 1/(1 - (r/2n)²

put 1/n = dx & r/2n = x

then integral limits → 0 & 1/2

hence....

S = $ (0, ½) dx / √(1-x²)

S = $ (0, ½) sin^(-1) x

S = sin^(-1) (1/2) - sin^(-1) (0)

S = π/6