JEE Advance - Mathematics (1998 - No. 27)
If $${a_n} = \sum\limits_{r = 0}^n {{1 \over {{}^n{C_r}}},\,\,\,then\,\,\,\sum\limits_{r = 0}^n {{r \over {{}^n{C_r}}}} } $$ equals
$$\left( {n - 1} \right){a_n}$$
$$n{a_n}$$
$${1 \over 2}n{a_n}$$
None of the above