`underset(nrarroo)"Lt"underset(2n+1)overset(3n)sum(n)/(r^(2)-n^(2))` is equal to :

Options

A. `ln sqrt((2)/(3))`
B. ln `sqrt((3)/(2))`
C. `"ln" (2)/(3)`
D. `"ln" (3)/(2)`

Answer

B

Explanation

We have,
`S=underset(n to oo)lim" " underset(r=2n+1)overset(3n)sum (n)/(r^(2)-n^(2))`
`rArr S=underset(n to oo)lim" " underset(r=2n+1)overset(3n)sum (1)/(((r)/(n))^(2)-1)*(1)/(n)`
`rArr S=underset(2)overset(3)int(1)/(x^(2)-1)dx=(1)/(2)[ log_(e)|(x-1)/(x+1)|]_(2)^(3)`
`rArr S=(1)/(2)[ "ln" (1)/(2)-"ln"(1)/(3)]=(1)/(2)ln ((3)/(2))= lnsqrt((3)/(2))`