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If S(n) = Sigma (r=0)^(n) 1//""^(n)C(r) ...

If `S_(n) = Sigma _(r=0)^(n) 1//""^(n)C_(r) and t_(n) = Sigma _(r=0)^(n) r//""^(n)C_(r),` then `t_(n)//S_(n)` is equal to

A

`(1)/(2) n`

B

`(1)/(2) n-1`

C

`n-1`

D

`(2n-1)/(2)`

Text Solution

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The correct Answer is:
A
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