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 B A is equal to

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 $\frac{B}{A}$ is equal to

Smart Achievers · Accepted Answer

(1)

$Now, A = {}^{2 n}C_{n}$ $and$ $B = {}^{2 n - 1}C_{n}$

$Thus, \frac{B}{A} = \frac{(2 n - 1)!}{n! (n - 1)!} \times \frac{n! n!}{(2 n)!} = \frac{n}{2 n} = \frac{1}{2}$

Solution

Q.
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 $\frac{B}{A}$ is equal to

1 $\frac{1}{2}$

2 $2$

3 $1$

4 $\frac{1}{n}$

Ans.
(1)

$Now, A = {}^{2 n}C_{n}$ $and$ $B = {}^{2 n - 1}C_{n}$

$Thus, \frac{B}{A} = \frac{(2 n - 1)!}{n! (n - 1)!} \times \frac{n! n!}{(2 n)!} = \frac{n}{2 n} = \frac{1}{2}$

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Solution

Q. If A and B are coefficients of xn in the expansions of (1+x)2n and (1+x)2n-1 respectively, then BA is equal to

1 12

2 2

3 1

4 1n

Ans. (1) Now, A=Cn2n and B=Cn2n-1 Thus, BA=(2n-1)!n!(n-1)!×n!n!(2n)!=n2n=12