First slide

Binomial theorem for positive integral Index

Question

$C_{0} C_{r} + C_{1} C_{r + 1} + C_{2} C_{r + 2} + \dots + C_{n - r} C_{n}$ is equal to

Moderate

A

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

$\frac{n!}{r! (n + r)!}$
C

$\frac{n!}{(n - r)!}$
D

None of these

Solution

We know,
and $\begin{array}{l} (1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{r} x^{r} + \dots \dots (i) \\ {(1 + \frac{1}{x})}^{n} = C_{0} + C_{1} \frac{1}{x} + C_{2} \frac{1}{x^{2}} + \dots + C_{r} \frac{1}{x^{r}} \end{array} + C_{r + 1} \frac{1}{x^{r + 1}} + C_{r + 2} \frac{1}{x^{r + 2}} \dots C_{n} \frac{1}{x^{n}} \dots (ii)$
on multiplying Eqs. (i) and (ii), equating coefficient of $x^{r} in \frac{1}{x^{n}} {(1 + x)}^{2 n} orthecoefficientof x^{n + r} in (1 + x)^{2 n},$
we get the value of required expression which is $^{2 n} C_{n + r} = \frac{(2 n)!}{(n - r)! (n + r)!}$

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