The sum of the coefficients of x^{2/3} and x^ | vedclass

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

Question

(A) The general term in the expansion of $(x^{2/3} + \frac{1}{2}x^{-2/5})^9$ is given by $T_{r+1} = {}^9C_r (x^{2/3})^{9-r} (\frac{1}{2} x^{-2/5})^r$.

Vedclass · Accepted Answer

$21/4$

Vedclass · Answer

(A) The general term in the expansion of $(x^{2/3} + \frac{1}{2}x^{-2/5})^9$ is given by $T_{r+1} = {}^9C_r (x^{2/3})^{9-r} (\frac{1}{2} x^{-2/5})^r$.$T_{r+1} = {}^9C_r (\frac{1}{2})^r x^{6 - \frac{2r}{3} - \frac{2r}{5}} = {}^9C_r (\frac{1}{2})^r x^{6 - \frac{16r}{15}}$.For the coefficient of $x^{2/3}$,set $6 - \frac{16r}{15} = \frac{2}{3}$.$6 - \frac{2}{3} = \frac{16r}{15}$ $\Rightarrow \frac{16}{3} = \frac{16r}{15}$ $\Rightarrow r = 5$.Coefficient of $x^{2/3} = {}^9C_5 (\frac{1}{2})^5 = 126 	imes \frac{1}{32} = \frac{63}{16}$.For the coefficient of $x^{-2/5}$,set $6 - \frac{16r}{15} = -\frac{2}{5}$.$6 + \frac{2}{5} = \frac{16r}{15}$ $\Rightarrow \frac{32}{5} = \frac{16r}{15}$ $\Rightarrow r = 6$.Coefficient of $x^{-2/5} = {}^9C_6 (\frac{1}{2})^6 = 84 	imes \frac{1}{64} = \frac{21}{16}$.Sum of coefficients $= \frac{63}{16} + \frac{21}{16} = \frac{84}{16} = \frac{21}{4}$.

The sum of the coefficients of $x^{2/3}$ and $x^{-2/5}$ in the binomial expansion of $(x^{2/3} + \frac{1}{2}x^{-2/5})^9$ is:

Similar Questions

The number of irrational terms in the expansion of $(3^{\frac{1}{8}}+5^{\frac{1}{4}})^{84}$ is

The coefficient of $x^5$ in the expansion of $(x + 3)^6$ is

The sum of the coefficients of the first three terms in the expansion of $(x - \frac{3}{x^2})^m$,where $x \neq 0$ and $m$ is a natural number,is $559$. Find the term of the expansion containing $x^3$. (in $x^3$)

If the $2^{\text{nd}}$,$3^{\text{rd}}$,and $4^{\text{th}}$ terms in the expansion of $(x+a)^{n}$ are $96, 216$,and $216$ respectively,and $n$ is a positive integer,then $a+x=$

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module