The coefficient of x^{-6} in the expansion of | vedclass

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

Question

(D) The general term $T_{r+1}$ in the expansion of $\left(\frac{4x}{5} + \frac{5}{2x^2}\right)^9$ is given by:$T_{r+1} = {}^9C_r \left(\frac{4x}{5}\ri

Vedclass · Accepted Answer

$5040$

Vedclass · Answer

(D) The general term $T_{r+1}$ in the expansion of $\left(\frac{4x}{5} + \frac{5}{2x^2}ight)^9$ is given by:$T_{r+1} = {}^9C_r \left(\frac{4x}{5}ight)^{9-r} \left(\frac{5}{2x^2}ight)^r$$= {}^9C_r \left(\frac{4}{5}ight)^{9-r} \left(\frac{5}{2}ight)^r x^{9-r} x^{-2r}$$= {}^9C_r \left(\frac{4}{5}ight)^{9-r} \left(\frac{5}{2}ight)^r x^{9-3r}$To find the coefficient of $x^{-6}$,we set the exponent of $x$ equal to $-6$:$9 - 3r = -6$$3r = 15 \Rightarrow r = 5$Substituting $r = 5$ into the expression for the coefficient:Coefficient $= {}^9C_5 \left(\frac{4}{5}ight)^{9-5} \left(\frac{5}{2}ight)^5$$= 126 	imes \left(\frac{4}{5}ight)^4 	imes \left(\frac{5}{2}ight)^5$$= 126 	imes \frac{256}{625} 	imes \frac{3125}{32}$$= 126 	imes 4 	imes 10 = 5040$

The coefficient of $x^{-6}$ in the expansion of $\left(\frac{4x}{5} + \frac{5}{2x^2}\right)^9$ is $........$.

Similar Questions

The coefficient of $x^2$ in the expansion of $(2x^2 + \frac{1}{x})^{10}, x \neq 0$,is:

The coefficients of three consecutive terms in the expansion of $(1+a)^{n}$ are in the ratio $1:7:42$. Find $n$.

The ratio of the coefficient of $x^{15}$ to the term independent of $x$ in the expansion of $(x^2 + \frac{2}{x})^{15}$ is

The numerically greatest term in the binomial expansion of $(2x - 3y)^5$ when $x = \frac{3}{2}$ and $y = \frac{2}{3}$ is

The square root of the independent term in the expansion of $\left(\frac{2x^2}{5} + \sqrt{\frac{5}{x}}\right)^{10}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module