Let us take this binomial expansion:
$\left ( 1 + x \right )^{^{n}} = \binom{n}{0} + \binom{n}{1} x + \binom{n}{2}x^{2} + \binom{n}{3}x^{3} + \cdot + \cdot + \binom{n}{n}x^{n}$
Multiplying it with x gives:
$x\left ( 1 + x \right )^{^{n}} = \binom{n}{0}x + \binom{n}{1} x^{2} + \binom{n}{2}x^{3} + \binom{n}{3}x^{4} + \cdot + \cdot + \binom{n}{n}x^{n+1}$
Now differentiate both sides w.r.t x:
$x\left [ n\left ( 1 + x \right )^{n-1} \right ]+ \left ( 1 + x \right )^{^{n}} = \binom{n}{0} + 2\binom{n}{1} x^{1} + 3\binom{n}{2}x^{2} + 4\binom{n}{3}x^{3} + \cdot + \cdot + \left ( n+1 \right )\binom{n}{n}x^{n}$
Now, substitute x = 1:
$\left [ n\left ( 2 \right )^{n-1} \right ]+ \left ( 2 \right )^{^{n}} = \binom{n}{0} + 2\binom{n}{1} + 3\binom{n}{2} + 4\binom{n}{3} + \cdot + \cdot + \left ( n+1 \right )\binom{n}{n}$
Thus, the answer is option (A).