First slide

Binomial theorem for positive integral Index

Question

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

Moderate

A

$A = B$
B

$2 A = B$
C

$A = 2 B$
D

none of these

Solution

We have ,
A = Coefficient of $x^{n}$ in $(1 + x)^{2 n}$
$\begin{array}{l} \Rightarrow A =^{2 n} C_{n} \\ \Rightarrow A = {\frac{2 n}{n}}^{2 n - 1} C_{n - 1} = 2^{2 n - 1} C_{n - 1} [∵^{n} C_{r} = {\frac{n}{r}}^{n - 1} C_{r - 1}] \end{array}$
and ,
B = Coefficient of $x^{n}$ in $(1 + x)^{2 n - 1}$
$\Rightarrow B =^{2 n - 1} C_{n} =^{2 n - 1} C_{n - 1}$ $[∵ n_{r} =^{n} C_{n - r}]$
Clearly, $A = 2 B$ .

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