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

A
STATEMENT-1 is True, STATEMENT-2 is True;
STATEMENT-2 is a correct explanation for STATEMENT-1
B
STATEMENT-1 is True, STATEMENT-2 is True;
STATEMENT-2 is NOT a correct explanation for STATEMENT-1
C
STATEMENT-1 is True, STATEMENT-2 is False
D
STATEMENT-1 is False, STATEMENT-2 is True

Solution:

$C_{0}^{2} + C_{1}^{2} + \dots + C_{n}^{2}$

$= C_{0} C_{n} + C_{1} C_{n - 1} + \dots + C_{n} C_{0}$

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

$=^{2 n} C_{n}$

$∴$ Statement-2 is true

Let $S = 5 C_{0}^{2} + 7 C_{1}^{2} + 9 C_{2}^{2} + \dots + (5 + 2 n) C_{n}^{2}$ (1)

Using

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

$S = (5 + 2 n) C_{0}^{2} + (3 + 2 n) C_{1}^{2} + \dots + 5 C_{n}^{2}$ (2)

Adding (1) and (2), $w e$ get

$\begin{array}{l} 2 S = (10 + 2 n) ({C_{0}}^{2} + {C_{1}}^{2} + \dots + {C_{n}}^{2}) \\ S = (5 + n) (^{2 n} C_{n}) = (5 + n) \frac{(2 n)}{n! n!} \end{array}$

Let $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$
Statement-1: $5 C_{0}^{2} + 7 C_{1}^{2} + 9 C_{2}^{2} + \dots + (5 + 2 n) C_{n}^{2}$
$= (5 + n) \frac{(2 n)!}{n! n!}$
Statement-2: $C_{0}^{2} + C_{1}^{2} + \dots + C_{n}^{2} =^{2 n} C_{n}$

Solution:

Related content