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If `S_n=sum_(r=0)^n 1/(nC_r) and sum_(r=0)^n r/(nC_r),` then `t_n/S_n=`A. `(2 n - 1)/(2) `B. `(n)/(2) -1`C. n-1D. `(n)/(2)` |
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Answer» Correct Answer - d We have, `s_(n) = sum_(r=0)^(n) (r)/(""^(n)C_(r)) = (n)/(2) s_(n)` [ See Example 37] `rArr (t_(n))/(s_(n)) = (n)/(2)` . |
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