Suppose A and B are the coefficients of 30^{t | vedclass

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

Question

(B) The $r^{	ext{th}}$ term in the expansion of $(1+x)^m$ is given by $T_r = {}^{m}C_{r-1} x^{r-1}$.For the expansion of $(1+x)^{2n-1}$:The $30^{	ex

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(B) The $r^{	ext{th}}$ term in the expansion of $(1+x)^m$ is given by $T_r = {}^{m}C_{r-1} x^{r-1}$.For the expansion of $(1+x)^{2n-1}$:The $30^{	ext{th}}$ term is $T_{30} = {}^{2n-1}C_{29} x^{29}$,so $A = {}^{2n-1}C_{29}$.The $12^{	ext{th}}$ term is $T_{12} = {}^{2n-1}C_{11} x^{11}$,so $B = {}^{2n-1}C_{11}$.Given $2A = 5B$,we have $2({}^{2n-1}C_{29}) = 5({}^{2n-1}C_{11})$.Using the property ${}^{n}C_r = {}^{n}C_{n-r}$,we know ${}^{2n-1}C_{29} = {}^{2n-1}C_{(2n-1)-29} = {}^{2n-1}C_{2n-30}$.Thus,$2({}^{2n-1}C_{2n-30}) = 5({}^{2n-1}C_{11})$.Using the ratio formula $\frac{{}^{n}C_r}{{}^{n}C_{r-1}} = \frac{n-r+1}{r}$,we can solve for $n$ by comparing the terms or testing the options.For $n=21$,$2n-1 = 41$.$2({}^{41}C_{29}) = 5({}^{41}C_{11})$.Since ${}^{41}C_{29} = {}^{41}C_{12}$,we check $2({}^{41}C_{12}) = 5({}^{41}C_{11})$.$2 	imes \frac{41!}{12! 29!} = 5 	imes \frac{41!}{11! 30!}$.$2 	imes \frac{1}{12} = 5 	imes \frac{1}{30} \implies \frac{2}{12} = \frac{5}{30} \implies \frac{1}{6} = \frac{1}{6}$.Thus,$n = 21$ is the correct answer.

Suppose $A$ and $B$ are the coefficients of $30^{\text{th}}$ and $12^{\text{th}}$ terms respectively in the binomial expansion of $(1+x)^{2n-1}$. If $2A = 5B$,then $n$ is equal to:

Similar Questions

$x^r$ occurs in the expansion of ${\left( x^3 + \frac{1}{x^4} \right)^n}$ provided -

The ratio of the $5^{th}$ term from the beginning to the $5^{th}$ term from the end in the binomial expansion of $\left( 2^{1/3} + \frac{1}{2(3)^{1/3}} \right)^{10}$ is

If the second,third and fourth terms in the expansion of $(x+y)^{n}$ are $135$,$30$ and $\frac{10}{3}$,respectively,then $6(n^3+x^2+y)$ is equal to .............

The sum of all possible values of $n \in N$ such that the coefficients of $x$, $x^2$, and $x^3$ in the expansion of $(1+x^2)^2(1+x)^n$ are in arithmetic progression is:

If $a, b,$ and $c$ are the greatest values of $^{19}C_{p}, ^{20}C_{q},$ and $^{21}C_{r}$ respectively,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module