Question

	Column - I		Column - II
(A)	${\sum }_{r=1}^{n}r{.}^n{C}_r$ is equal to	(P)	$(n+2){.2}^{n-1}$
(B)	${\sum }_{r=1}^{n+1}r{.}^n{C}_{r-1}$ is equal to	(Q)	$(n+1){.}^{2n}{C}_n$
(C)	${\sum }_{r=0}^n(2r+1){.}^n{C}_r$ is equal to	(R)	$(n+1){.2}^n$
(D)	${\sum }_{r=0}^n(2r+1).{(}^n{C}_r{)}^2$ is equal to	(S)	$n{.2}^{n-1}$

Answer 1

$(1+x{)}^n{=}^n{C}_0{+}^n{C}_{1}x+.....................{+}^n{C}_n{x}^n$

Differentiating,

$n(1+x{)}^{n-1}{=}^n{C}_1+{2.}^n{C}_{2}x+.......................n{.}^n{C}_n{x}^n$

Putting x = 1 we get required sum = $n{.2}^{n-1}$

$x(1+x{)}^n{=}^n{C}_{0}x{+}^n{C}_1{x}^2+.....................{+}^n{C}_n{x}^{n+1}$

Differentiating,

$n(1+x{)}^{n-1}x+(1+x{)}^n=$${}^n{C}_0+{2.}^n{C}_{1}x+{3.}^n{C}_2{x}^2+.......................(n+1){.}^n{C}_n{x}^n$

Putting x =1. we get required sum:

$S=n{.2}^{n-1}+2^n=2^{n-1}(n+2)$

$(1+x^2{)}^n.x{=}^n{C}_{0}x{+}^n{C}_1{x}^3+.....................{+}^n{C}_n{x}^{2n+1}$

Differentiating,

$n(1+x^2{)}^{n-1}2x^2+(1+x^2{)}^n=$${}^n{C}_0+{3.}^n{C}_1{x}^2+{5.}^n{C}_2{x}^4+.......................(2n+1){.}^n{C}_n{x}^{2n}$

Putting x =1. we get required sum:

$S=n{.2}^{n-1}2+2^n=2^n(n+1)$

Obviously the last answer will be the only option remaining.