Let (1+x)n=C0+C1x+C2x2+⋯+CnxnStatement-1: 5C02+7C12+9C22+⋯+(5+2n)Cn2=(5+n)(2n)!n!n!Statement-2: C02+C12+⋯+Cn2=2nCn

# Let $\left(1+x{\right)}^{n}={C}_{0}+{C}_{1}x+{C}_{2}{x}^{2}+\cdots +{C}_{n}{x}^{n}$Statement-1: $5{C}_{0}^{2}+7{C}_{1}^{2}+9{C}_{2}^{2}+\cdots +\left(5+2n\right){C}_{n}^{2}$$=\left(5+n\right)\frac{\left(2n\right)!}{n!n!}$Statement-2: ${C}_{0}^{2}+{C}_{1}^{2}+\cdots +{C}_{n}^{2}{=}^{2n}{C}_{n}$

1. A

STATEMENT-1 is True, STATEMENT-2 is True;
STATEMENT-2 is a correct explanation for STATEMENT-1

2. B

STATEMENT-1 is True, STATEMENT-2 is True;
STATEMENT-2 is NOT a correct explanation for STATEMENT-1

3. C

STATEMENT-1 is True, STATEMENT-2 is False

4. D

STATEMENT-1 is False, STATEMENT-2 is True

Register to Get Free Mock Test and Study Material

+91

Live ClassesRecorded ClassesTest SeriesSelf Learning

Verify OTP Code (required)

### Solution:

${C}_{0}^{2}+{C}_{1}^{2}+\dots +{C}_{n}^{2}$

$={C}_{0}{C}_{n}+{C}_{1}{C}_{n-1}+\cdots +{C}_{n}{C}_{0}$

= number of ways of choosing n persons out of
n men and n women

${=}^{2n}{C}_{n}$

$\therefore$ Statement-2 is true

Let  $S=5{C}_{0}^{2}+7{C}_{1}^{2}+9{C}_{2}^{2}+\cdots +\left(5+2n\right){C}_{n}^{2}$              (1)

Using

${C}_{r}={C}_{n-r}$ we can rewrite (1) as

$S=\left(5+2n\right){C}_{0}^{2}+\left(3+2n\right){C}_{1}^{2}+\dots +5{C}_{n}^{2}$                    (2)

Adding (1) and (2),  $we$ get  