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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