In the binomial expansion of (1+x)^{2k},if it | vedclass

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

Question

(B) The middle term in the binomial expansion of $(1+x)^n$,when $n$ is even,is $T_{r+1} = {}^{n}C_r x^r$,where $r = \frac{n}{2}$.For $(1+x)^{2k}$,the

Vedclass · Accepted Answer

$(-\frac{k+1}{k}, \frac{k+1}{k})$

Vedclass · Answer

(B) The middle term in the binomial expansion of $(1+x)^n$,when $n$ is even,is $T_{r+1} = {}^{n}C_r x^r$,where $r = \frac{n}{2}$.For $(1+x)^{2k}$,the middle term is $T_{k+1} = {}^{2k}C_k x^k$.Given that the middle term is the only numerically greatest term,we must have $|T_{k+1}| > |T_k|$ and $|T_{k+1}| > |T_{k+2}|$.First,$|\frac{T_{k+1}}{T_k}| > 1$ $\Rightarrow |\frac{{}^{2k}C_k x^k}{{}^{2k}C_{k-1} x^{k-1}}| > 1$ $\Rightarrow |\frac{k+1}{k} x| > 1$ $\Rightarrow |x| > \frac{k}{k+1}$.Second,$|\frac{T_{k+1}}{T_{k+2}}| > 1$ $\Rightarrow |\frac{{}^{2k}C_k x^k}{{}^{2k}C_{k+1} x^{k+1}}| > 1$ $\Rightarrow |\frac{k+1}{k x}| > 1$ $\Rightarrow |x| Combining these,we get $\frac{k}{k+1} However,for the middle term to be the *only* greatest term,the condition simplifies to $|x| < \frac{k+1}{k}$ (specifically,the range of $x$ for which the middle term is the greatest is $(-\frac{k+1}{k}, \frac{k+1}{k})$).

In the binomial expansion of $(1+x)^{2k}$,if its middle term is the only numerically greatest term,then $x$ lies in the interval

Similar Questions

If the coefficients of $x^2$ and $x^3$ in the expansion of $(3+ax)^9$ are the same, then the value of $a$ is

Find the coefficient of $x^{5}$ in the expansion of $(x+3)^{8}$.

The term independent of $x$ in ${\left( {\sqrt x - \frac{2}{x}} \right)^{18}}$ is

The coefficient of $x^{18}$ in the expansion of $(x^4-\frac{1}{x^3})^{15}$ is $...........$.

In the expansion of ${\left( \frac{a}{x} + bx \right)^{12}}$,the coefficient of $x^{-10}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module