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If a(n) =sum(r=0)^(n) (1)/(""^(n)C(r ))...

If `a_(n) =sum_(r=0)^(n) (1)/(""^(n)C_(r ))`, then `sum_(r=0)^(n) (r )/(""^(n)C_(r ))` equals

A

`(n - 1) a_(n)`

B

`n a_(n)`

C

`(1)/(2) n a_(n)`

D

None of these

Text Solution

Verified by Experts

Let `b=sum _(r=0)^(n)(r)/(.^Nc_r)=sum_(r=0)^(n)(-(n-r))/(.^Nc_R)`
`=n sum_(r=0)^(n)(1)/(.^n C_r)-sum_(r=0)^(n)(n-r)/(.^C_r)`
`=na_(n)-sum_(r=0)^(n) (n-r)/(.^nC_(n-r))[because .^Nc_r=.^nC_(n-r)]`
`=na_(n)-brArr 2b=na_(n)rArr b=(n)/(2)a_(n)`.
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