If the 7^{th} term from the beginning in the | vedclass

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

Question

(B) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = {^nC_r} a^{n-r} b^r$.For the $7^{th}$ term,$r = 6$ and $n = 9$.$T_

Vedclass · Accepted Answer

$e$

Vedclass · Answer

(B) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = {^nC_r} a^{n-r} b^r$.For the $7^{th}$ term,$r = 6$ and $n = 9$.$T_7 = {^9C_6} \left( \frac{3}{(84)^{1/3}} ight)^{9-6} (\sqrt{3} \ln x)^6 = 729$.$T_7 = {^9C_3} \left( \frac{3^3}{84} ight) (3^3) (\ln x)^6 = 729$.$T_7 = 84 	imes \frac{27}{84} 	imes 27 	imes (\ln x)^6 = 729$.$27 	imes 27 	imes (\ln x)^6 = 729$.$729 	imes (\ln x)^6 = 729$.$(\ln x)^6 = 1$.$\ln x = \pm 1$.If $\ln x = 1$,then $x = e$.If $\ln x = -1$,then $x = \frac{1}{e}$.Given the options,$x = e$ is the correct value.

If the $7^{th}$ term from the beginning in the binomial expansion of ${\left( {\frac{3}{{{{\left( {84} \right)}^{\frac{1}{3}}}}} + \sqrt 3 \ln x} \right)^9}$ for $x > 0$ is equal to $729$,then the possible value of $x$ is:

Similar Questions

In the expansion of ${\left( {3x - \frac{1}{{{x^2}}}} \right)^{10}}$,the $5^{th}$ term from the end is:

If $A$ and $B$ are the coefficients of $x^n$ in the expansions of $(1 + x)^{2n}$ and $(1 + x)^{2n - 1}$ respectively,then

If the number of integral terms in the expansion of $(3^{\frac{1}{2}} + 5^{\frac{1}{8}})^n$ is exactly $33$,then the least value of $n$ is

The coefficient of the middle term in the expansion of $(1 + x)^{10}$ is

If the coefficients of $x^{-2}$ and $x^{-4}$ in the expansion of ${\left( {{x^{\frac{1}{3}}} + \frac{1}{{2{x^{\frac{1}{3}}}}}} \right)^{18}}, (x > 0),$ are $m$ and $n$ respectively,then $\frac{m}{n}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module