Question

nC1 + 2. n C2 + 3. n C3 +-----+ n. n Cn =_____

(a) n. 2n-1

(b) (n-1) 2n-1

(c) (n+1) 2n-1

(d) (n-1) 2n

Answer 1

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 + 2nC2 + 3nC3 + ---- n ∙ nCn = n ∙ 2n–1