First slide

Binomial theorem for positive integral Index

Question

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ , then $C_{0} C_{2} + C_{1} C_{3} + C_{2} C_{4} + \dots + C_{n - 2} C_{n} =$

Moderate

A

$\frac{(2 n)!}{(n!)^{2}}$
B

$\frac{(2 n)!}{(n - 1)! (n + 1)!}$
C

$\frac{(2 n)!}{(n - 2)! (n + 2)!}$
D

$\frac{(2 n)!}{({(n - 1)!)}^{2}}$

Solution

$\begin{matrix} (1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + C_{3} x^{3} + \dots + C_{n - 1} x^{n - 1} + C_{n} x^{n} (1) \\ (x + 1)^{n} = C_{0} x^{n} + C_{1} x^{n - 1} + C_{2} x^{n - 2} + \dots + C_{n - 1} x + C_{n} (2) \end{matrix}$
Multiplying Eqs. (1) and (2) and equating the coefficient of $x^{n - 2}$ , we get
$\begin{matrix} C_{0} C_{2} + C_{1} C_{3} + C_{2} C_{4} + \dots + C_{n - 2} C_{n} \\ = Coefficient of x^{n - 2} in (1 + x)^{2 n} \\ =^{2 n} C_{n - 2} \\ = \frac{(2 n)!}{(n - 2)! (n + 2)!} \end{matrix}$

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