Find the sum of the series:
`(1)/((1xx3))+(1)/((3xx5))+(1)/((5xx7))+...+(1)/((2n-1)(2n+1))`

Answer

`(n)/((2n+1))`

Explanation

`T_(n)=(1)/((2n-1)(2n+1))=(A)/(2n-1)+(B)/(2n+1)=(A(2n+1)+B(2n-1))/((2n-1)(2n+1))=(2(A+B)n+(A-B))/((2n-1)(2n+1))`
`therefore 2(A+B)=0 and A-B=1 rArr A+B=0, A-B=1 rArr A=(1)/(2), B=(-1)/(2).`
`therefore T_(n)=(1)/(2)xx{(1)/(2n-1)-(1)/(2n+1)}`.
`therefore T_(1)=(1)/(2)xx((1)/(1)-(1)/(3))`
`T_(2)=(1)/(2)xx((1)/(3)-(1)/(5))`
`T_(3)=(1)/(2)xx((1)/(5)-(1)/(7))`
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`T_(n-1)=(1)/(2)xx{(1)/(2n-3)-(1)/(2n-1)}`
`T_(n)=(1)/(2)xx{(1)/(2n-1)-(1)/(2n+1)}`.
Adding columnwise, we get
`S_(n)=(T_(1)+T_(2)+T_(3)+... +T_(n))=(1)/(2)(1-(1)/(2n+1))=(2n+1-1)/(2(2n+1))=(n)/((2n+1)).`