The interval in which x must lie so that the | vedclass

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

Question

(B) The greatest coefficient in the expansion of $(1 + x)^{2n}$ is the middle term coefficient,which is $^{2n}C_n$.For the term $T_{n+1} = ^{2n}C_n x^

Vedclass · Accepted Answer

$\left( \frac{n}{n + 1}, \frac{n + 1}{n} \right)$

Vedclass · Answer

(B) The greatest coefficient in the expansion of $(1 + x)^{2n}$ is the middle term coefficient,which is $^{2n}C_n$.For the term $T_{n+1} = ^{2n}C_n x^n$ to be the greatest term,it must satisfy $T_{n+1} \ge T_n$ and $T_{n+1} \ge T_{n+2}$.First,$T_{n+1} \ge T_n \Rightarrow ^{2n}C_n x^n \ge ^{2n}C_{n-1} x^{n-1}$.This simplifies to $x \ge \frac{^{2n}C_{n-1}}{^{2n}C_n} = \frac{n}{n+1}$.Second,$T_{n+1} \ge T_{n+2} \Rightarrow ^{2n}C_n x^n \ge ^{2n}C_{n+1} x^{n+1}$.This simplifies to $x \le \frac{^{2n}C_n}{^{2n}C_{n+1}} = \frac{n+1}{n}$.Thus,the required interval is $\left( \frac{n}{n + 1}, \frac{n + 1}{n} \right)$.

The interval in which $x$ must lie so that the greatest term in the expansion of $(1 + x)^{2n}$ has the greatest coefficient,is

Similar Questions

The term independent of $x$ in the expansion of ${\left( {2x - \frac{3}{x}} \right)^6}$ is

The number of irrational terms in the expansion of $(5^{1/2} + 7^{1/8})^{1024} + (5^{1/2} - 7^{1/8})^{1024}$ is

The middle term in the expansion of ${\left( {1 - \frac{1}{x}} \right)^n}{\left( {1 - x} \right)^n}$ in powers of $x$ is

In the binomial $(2^{1/3} + 3^{-1/3})^n$,if the ratio of the seventh term from the beginning of the expansion to the seventh term from its end is $1/6$,then $n =$

If the coefficients of $r$ th and $(r+1)$ th terms in the expansion of $(3+7x)^{29}$ are equal,then $r$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module