Gujarati
Hindi
7.Binomial Theorem
normal

If $S_n =$$\sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} $ and $T_n =$$\sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $ then $\frac{{{T_n}}}{{{S_n}}}$ is equal to

A

$\frac{n}{2}$

B

$\frac{n}{2} - 1$

C

$n - 1$

D

$\frac{{2n - 1}}{2}$

Solution

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Standard 11
Mathematics

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