Find the sum to `n` terms of the series: `1/(1. 3)+1/(3. 5)+1/(5. 7)+`

Options

A. `(1)/(2n+1)`
B. `(2n)/(2n+1)`
C. `(n)/(2n+1)`
D. `(2n)/(n+1)`

Answer

C

Explanation

Let `T_(r)" be the "r^(th)` term of the given series. Then,
`T_(r)=(1)/((2r-1)(2r+1)),r=1,2,3, . . .n`
`rArr" "T_(r)=(1)/(2)((1)/(2r-1)-(1)/(2r+1))`
Let S be the required sum. Then,
`S=underset(r-1)overset(n)sumT_(r)`
`S=(1)/(2){:[((1)/(1)-(1)/(3))+((1)/(3)-(1)/(5))+((1)/(5)-(1)/(7))+ . . . +((1)/(2n-1)-(1)/(2n+1))]:}`
`rArr" "S=(1)/(2){1-(1)/(2n+1)}=(n)/(2n+1)`