The value of the numerically greatest term in | vedclass

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

Question

(A) For the expansion of $(a + b)^n$,the $(r+1)^{	ext{th}}$ term is given by $T_{r+1} = {}^{n}C_r a^{n-r} b^r$. Given $(2x + 3y)^{11}$,we substitute

Vedclass · Accepted Answer

$462$

Vedclass · Answer

(A) For the expansion of $(a + b)^n$,the $(r+1)^{	ext{th}}$ term is given by $T_{r+1} = {}^{n}C_r a^{n-r} b^r$. Given $(2x + 3y)^{11}$,we substitute $x = \frac{1}{2}$ and $y = \frac{1}{3}$,so $a = 2(\frac{1}{2}) = 1$ and $b = 3(\frac{1}{3}) = 1$. The expansion becomes $(1 + 1)^{11}$. The general term is $T_{r+1} = {}^{11}C_r (1)^{11-r} (1)^r = {}^{11}C_r$. To find the greatest term,we look for the maximum value of ${}^{11}C_r$ for $r = 0, 1, \dots, 11$. The binomial coefficients ${}^{n}C_r$ are maximum when $r = \frac{n}{2}$ if $n$ is even,or $r = \frac{n-1}{2}$ and $r = \frac{n+1}{2}$ if $n$ is odd. Here $n = 11$ (odd),so the maximum values occur at $r = \frac{11-1}{2} = 5$ and $r = \frac{11+1}{2} = 6$. Thus,the greatest term is ${}^{11}C_5 = {}^{11}C_6 = \frac{11 	imes 10 	imes 9 	imes 8 	imes 7}{5 	imes 4 	imes 3 	imes 2 	imes 1} = 462$.

The value of the numerically greatest term in the expansion of $(2x + 3y)^{11}$ when $x = \frac{1}{2}$ and $y = \frac{1}{3}$ is

Similar Questions

The coefficient of $x^5$ in the expansion of $(1+x^2)^5(1+x)^4$ is:

If the coefficients of $r$-th and $(r+1)$-th terms in the expansion of $(1+x)^{24}$ are in the ratio $12:13$,then $r$ is the root of the quadratic equation

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 $(x + 3)^6$ is

If the coefficients of the $r^{th}$ term and the $(r + 4)^{th}$ term are equal in the expansion of $(1 + x)^{20}$,then the value of $r$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module