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If `a_(n) =sum_(r=0)^(n) (1)/(""^(n)C_(r ))`, then `sum_(r=0)^(n) (r )/(""^(n)C_(r ))` equalsA. `(n - 1) a_(n)`B. `n a_(n)`C. `(1)/(2) n a_(n)`D. None of these |
Answer» 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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