If A and B are coefficients of x^{n} in the e | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The coefficient of $x^{n}$ in the expansion of $(1+x)^{2n}$ is given by $A = {}^{2n}C_{n}$. The coefficient of $x^{n}$ in the expansion of $(1+x)^

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The coefficient of $x^{n}$ in the expansion of $(1+x)^{2n}$ is given by $A = {}^{2n}C_{n}$. The coefficient of $x^{n}$ in the expansion of $(1+x)^{2n-1}$ is given by $B = {}^{2n-1}C_{n}$. Now, we calculate the ratio $A / B$: $\frac{A}{B} = \frac{{}^{2n}C_{n}}{{}^{2n-1}C_{n}}$ Using the formula ${}^{n}C_{r} = \frac{n}{r} 	imes {}^{n-1}C_{r-1}$, we have ${}^{2n}C_{n} = \frac{2n}{n} 	imes {}^{2n-1}C_{n-1}$. However, a simpler approach is: $\frac{A}{B} = \frac{\frac{(2n)!}{n!n!}}{\frac{(2n-1)!}{n!(n-1)!}} = \frac{(2n)!}{n!n!} 	imes \frac{n!(n-1)!}{(2n-1)!} = \frac{2n 	imes (2n-1)!}{n 	imes (n-1)! 	imes n!} 	imes \frac{n!(n-1)!}{(2n-1)!} = \frac{2n}{n} = 2$.

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

Similar Questions

If $(1-x+x^2)^{10}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$,then $2 a_2+3 a_3+4 a_4+\ldots+20 a_{20}=$

The coefficient of $x^5$ in the expansion of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$ is

The coefficient of $x^{4}$ in the expansion of $(1+x+x^{2}+x^{3})^{6}$ in powers of $x$ is:

The coefficient of $x^5$ in the expansion of $(x + 3)^6$ is

If $n$ is even,then in the expansion of ${\left( {1 + \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} + \dots} \right)^2}$,the coefficient of ${x^n}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module