If the coefficient of x^3 in the binomial exp | vedclass

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

Question

(D) The general term in the expansion of $(2 \sqrt{3} x^2 + \frac{1}{kx})^{12}$ is $T_{r+1} = {}^{12}C_r (2 \sqrt{3} x^2)^{12-r} (\frac{1}{kx})^r$.The

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(D) The general term in the expansion of $(2 \sqrt{3} x^2 + \frac{1}{kx})^{12}$ is $T_{r+1} = {}^{12}C_r (2 \sqrt{3} x^2)^{12-r} (\frac{1}{kx})^r$.The expression is $x^3 	imes T_{r+1} = {}^{12}C_r (2 \sqrt{3})^{12-r} (\frac{1}{k})^r x^{3 + 2(12-r) - r} = {}^{12}C_r (2 \sqrt{3})^{12-r} (\frac{1}{k})^r x^{27-3r}$.For the coefficient of $x^3$,we set $27 - 3r = 3$,which gives $3r = 24$,so $r = 8$.The coefficient is ${}^{12}C_8 (2 \sqrt{3})^{12-8} (\frac{1}{k})^8 = {}^{12}C_4 (2 \sqrt{3})^4 (\frac{1}{k^8}) = 880$.${}^{12}C_4 = \frac{12 	imes 11 	imes 10 	imes 9}{4 	imes 3 	imes 2 	imes 1} = 495$.$(2 \sqrt{3})^4 = 16 	imes 9 = 144$.So,$495 	imes 144 	imes \frac{1}{k^8} = 880$.$\frac{71280}{k^8} = 880 \Rightarrow k^8 = \frac{71280}{880} = 81$.$k^8 = 3^4$ $\Rightarrow k^2 = \sqrt{3}$ $\Rightarrow k = \sqrt[4]{3}$ is incorrect based on the provided options; re-evaluating: $k^8 = 81 = 3^4$,so $k = (3^4)^{1/8} = 3^{1/2} = \sqrt{3}$.

If the coefficient of $x^3$ in the binomial expansion of $x^3(2 \sqrt{3} x^2 + \frac{1}{kx})^{12}$ is $880$,then $k$ is equal to

Similar Questions

If the coefficients of $r$-th and $(r+1)$-th terms in the expansion of $(3+7x)^{29}$ are equal,then $r$ is equal to

The number of all possible values of $k$ for which the expansion $(\sqrt{x}+\sqrt[k]{y})^{10}$ will have exactly nine irrational terms is

The coefficient of the middle term in the binomial expansion in powers of $x$ of $(1 + \alpha x)^4$ and of $(1 - \alpha x)^6$ is the same if $\alpha$ equals

If the term independent of $x$ in the expansion of $\left(\sqrt{x}-\frac{k}{x^2}\right)^{10}$ is $405$,then $k=$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module