Correct option: (b) (n+1)2n
Explanation:
nc0 + 3nc1 + 5nc2 + ----- (2n + 1)ncn
= (nc0 + nc1 + ---- ncn) + 2 ∙ nc1 + 4nc2 + -----2n ∙ ncn
= 2n + (2nc1 + 4nc1 + ---- 2n ∙ ncn)
= 2n + 2(nc1 + 2nc2 + ---- n ∙ ncn)
= 2n + 2[n + {{2 ∙ n(n – 1)} / (2!)} + {{3 ∙ n(n – 1)(n – 2)} / (3!)} + --- n]
= 2n + 2[n + n(n – 1) + {{n(n – 1)(n – 2)} / (2!)} + --- n]
= 2n + 2n[1 + (n – 1) + {{(n – 1)(n – 2)} / (2!)} + --- 1]
= 2n + 2n(n–1c0 + n–1c1 + n–1c2 + ----- n–1cn–1)
= 2n + 2n ∙ 2n–1
= 2n + n ∙ 2n
= 2n(n + 1)
