In the expansion of left( frac{x}{2} - frac{3 | 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 $\left( \frac{x}{2} - \frac{3}{x^2} \right)^{10}$ is given by $T_{r+1} = {}^{10}C_r \left( \frac{x}{2} \right

Vedclass · Accepted Answer

$\frac{405}{256}$

Vedclass · Answer

(A) The general term in the expansion of $\left( \frac{x}{2} - \frac{3}{x^2} ight)^{10}$ is given by $T_{r+1} = {}^{10}C_r \left( \frac{x}{2} ight)^{10-r} \left( -\frac{3}{x^2} ight)^r$. Simplifying this,we get $T_{r+1} = {}^{10}C_r (-1)^r \frac{3^r}{2^{10-r}} x^{10-r-2r} = {}^{10}C_r (-1)^r \frac{3^r}{2^{10-r}} x^{10-3r}$. To find the coefficient of $x^4$,we set the exponent of $x$ equal to $4$: $10 - 3r = 4$ $\Rightarrow 3r = 6$ $\Rightarrow r = 2$. Substituting $r = 2$ into the expression for the term: $T_{2+1} = {}^{10}C_2 (-1)^2 \frac{3^2}{2^{10-2}} x^4 = \frac{10 	imes 9}{2 	imes 1} 	imes 1 	imes \frac{9}{2^8} x^4$. $T_3 = 45 	imes \frac{9}{256} x^4 = \frac{405}{256} x^4$. Thus,the coefficient of $x^4$ is $\frac{405}{256}$.

In the expansion of $\left( \frac{x}{2} - \frac{3}{x^2} \right)^{10}$,the coefficient of $x^4$ is

Similar Questions

Find the coefficient of $x^{6} y^{3}$ in the expansion of $(x+2 y)^{9}$.

The natural number $m$,for which the coefficient of $x$ in the binomial expansion of $\left( x^{m} + \frac{1}{x^{2}} \right)^{22}$ is $1540$,is

If the sum of the coefficients of $x^7$ and $x^{14}$ in the expansion of $\left(\frac{1}{x^3} - x^4\right)^n, x \neq 0$,is zero,then the value of $n$ is . . . . . . .

If the coefficients of $x^{7}$ and $x^{8}$ in the expansion of $(2+\frac{x}{3})^{n}$ are equal,then the value of $n$ is equal to $.....$

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module