The term independent of x in the expansion of | vedclass

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

Question

(B) The general term $T_{r+1}$ in the expansion of $\left(x-\frac{2}{\sqrt{x}}\right)^{21}$ is given by: $T_{r+1} = {}^{21}C_r (x)^{21-r} \left(-\frac

Vedclass · Accepted Answer

${}^{21}C_{14} 2^{14}$

Vedclass · Answer

(B) The general term $T_{r+1}$ in the expansion of $\left(x-\frac{2}{\sqrt{x}}\right)^{21}$ is given by: $T_{r+1} = {}^{21}C_r (x)^{21-r} \left(-\frac{2}{x^{1/2}}\right)^r$ $T_{r+1} = {}^{21}C_r (x)^{21-r} (-2)^r (x)^{-r/2}$ $T_{r+1} = {}^{21}C_r (-2)^r (x)^{21-r-r/2}$ $T_{r+1} = {}^{21}C_r (-2)^r (x)^{21-3r/2}$ For the term to be independent of $x$,the exponent of $x$ must be $0$: $21 - \frac{3r}{2} = 0$ $42 - 3r = 0$ $\Rightarrow 3r = 42$ $\Rightarrow r = 14$ Substituting $r = 14$ into the expression: $T_{14+1} = {}^{21}C_{14} (-2)^{14} = {}^{21}C_{14} 2^{14}$ Since ${}^{21}C_{14} = {}^{21}C_7$,the term is ${}^{21}C_{14} 2^{14}$.

The term independent of $x$ in the expansion of $\left(x-\frac{2}{\sqrt{x}}\right)^{21}$ is

Similar Questions

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

The coefficient of $x^n$ in the expansion of $(1 - 2x + 3x^2 - 4x^3 + \dots)^{-n}$ is

If the fourth term in the binomial expansion of $\left(\sqrt{\frac{1}{x^{1+\log _{10} x}}}+x^{\frac{1}{12}}\right)^{6}$ is equal to $200$,and $x > 1$,then the value of $x$ is

The constant term in the expansion of $\left(1+\frac{1}{x}\right)^{20}\left(30 x(1+x)^{29}+(1+x)^{30}\right)$ is

The sum of the real values of $x$ for which the middle term in the binomial expansion of ${\left( {\frac{{{x^3}}}{3} + \frac{3}{x}} \right)^8}$ equals $5670$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module