Question
$$1 \cdot {\,^n}{C_1} + 2 \cdot {\,^n}{C_2} + 3 \cdot {\,^n}{C_3} + ..... + n \cdot {\,^n}{C_n}$$ is equal to
A.
$$\frac{{n\left( {n + 1} \right)}}{4} \cdot {2^n}$$
B.
$${2^{n + 1}} - 3$$
C.
$$n \cdot {2^{n - 1}}$$
D.
None of these
Answer :
$$n \cdot {2^{n - 1}}$$
Solution :
$$r \cdot {\,^n}{C_r} = n \cdot {\,^{n - 1}}{C_{r - 1}}.$$
∴ the sum $$ = n\left\{ {^{n - 1}{C_0} + {\,^{n - 1}}{C_1} + {\,^{n - 1}}{C_2} + ..... + {\,^{n - 1}}{C_{n - 1}}} \right\}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = n \cdot {2^{n - 1}}.$$