The coefficient of the term independent of x | vedclass

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

Question

(A) The $(r+1)^{th}$ term in the expansion of $\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^{2}}\right)^{10}$ is given by$T_{r+1} = {^{10}C_r} \left(\sqrt{\f

Vedclass · Accepted Answer

$5/4$

Vedclass · Answer

(A) The $(r+1)^{th}$ term in the expansion of $\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^{2}}ight)^{10}$ is given by$T_{r+1} = {^{10}C_r} \left(\sqrt{\frac{x}{3}}ight)^{10-r} \left(\frac{3}{2x^2}ight)^r$$= {^{10}C_r} \frac{x^{(10-r)/2}}{3^{(10-r)/2}} \cdot \frac{3^r}{2^r x^{2r}}$$= {^{10}C_r} \frac{3^{r - (10-r)/2}}{2^r} x^{(10-r)/2 - 2r}$For the term to be independent of $x$,the exponent of $x$ must be zero:$\frac{10-r}{2} - 2r = 0 \implies 10 - r - 4r = 0 \implies 5r = 10 \implies r = 2$Substituting $r=2$ into the coefficient expression:Coefficient $= {^{10}C_2} \cdot \frac{3^{2 - (10-2)/2}}{2^2} = {^{10}C_2} \cdot \frac{3^{2-4}}{4} = \frac{10 	imes 9}{2} \cdot \frac{3^{-2}}{4} = 45 \cdot \frac{1}{9 	imes 4} = \frac{45}{36} = \frac{5}{4}$

The coefficient of the term independent of $x$ in the expansion of ${\left( {\sqrt {\frac{x}{3}} + \frac{3}{{2{x^2}}}} \right)^{10}}$ is

Similar Questions

Find the coefficient of $x^{15}$ in the product $(1 - x) (1 - 2x) (1 - 2^2 x) (1 - 2^3 x) \dots (1 - 2^{15} x)$.

The numerically greatest term in the expansion of $(x+3y)^{13}$, when $x=\frac{1}{2}$ and $y=\frac{1}{3}$ is:

If the third term in the binomial expansion of $(1 + x^{\log_2 x})^5$ equals $2560$,then a possible value of $x$ is

If the second term of the expansion $\left[ a^{\frac{1}{13}} + \frac{a}{\sqrt{a^{-1}}} \right]^n$ is $14a^{5/2}$,then the value of $\frac{^nC_3}{^nC_2}$ is:

The sum of all rational terms in the expansion of $(2+\sqrt{3})^8$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module