First slide

Binomial theorem for positive integral Index

Question

If $A$ and $B$ are coefficients of $x^{n}$ in the expansions of $(1 + x)^{2 n}$ and $(1 + x)^{2 n - 1}$ respectively, then

Moderate

A

$A = B$
B

$A = 2 B$
C

$2 A = B$
D

none of these

Solution

We know that coefficient of $x^{r}$ in the expansion of $(1 + x)^{m}$ is $^{m} C_{r}$ .
Thus, $A =^{2 n} C_{n}$ and $B =^{2 n - 1} C_{n^{}}$
We have $\frac{A}{B} = \frac{^{2 n} C_{n}}{^{2 n - 1} C_{n}} = \frac{(2 n)!}{n! n!} \frac{(n!) (n - 1)!}{(2 n - 1)!} = \frac{2 n}{n} = 2$
$\Rightarrow A = 2 B$

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