Correct option: (a) n. 2n-1
Explanation:
If a0, a1, a2 --- an are in A. P. then
a0 nC0 + a1 nC1 + a2 nC2 + ---- an nCn = (a0 + an) ∙ 2n–1
now
if a0, a1, a2 --- an = 0, 1, 2. 3 ----- n then
we can write,
0 + nC1 + 2 ∙ nC2 + 3 ∙ nC3 + ---- n ∙ nCn = (0 + n) ∙ 2n–1 = n ∙ 2n–1
∴ nC1 + 2 nC2 + 3 nC3 + ---- n ∙ nCn = n ∙ 2n–1
