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  • If in the expansion of <img alt="\small \left(2^x+\frac{1}{4^x}\right)^n, \frac{T_3}{T_2}=7" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Csmall%20%5Cleft%282%5Ex&plus;%5Cfrac%7B1%7D%

If in the expansion of \small \left(2^x+\frac{1}{4^x}\right)^n, \frac{T_3}{T_2}=7 and the sum of the coefficients of 2nd and 3rd terms is 36 , then the value of \small x is ...

Option: 1

x=\frac{1}{3}


Option: 2

x=-\frac{1}{3}


Option: 3

x=\frac{1}{2}


Option: 4

x=1


Answers (1)

best_answer

\begin{aligned} & \because \frac{T_3}{T_2}=7 \\ & \Rightarrow \quad \frac{{ }^n C_2\left(2^x\right)^{n-2}\left(\frac{1}{4^x}\right)^2}{{ }^n C_1\left(2^x\right)^{n-1}\left(\frac{1}{4^x}\right)^1}=7 \\ & \Rightarrow \quad \frac{{ }^n C_2 \cdot\left(2^x\right)^{n-6}}{{ }^n C_1\left(2^x\right)^{n-3}}=7 \end{aligned}

\Rightarrow \quad \frac{(n-1)}{2} \cdot\left(2^x\right)^{-3}=7                   ...(1)

Also given,{ }^n C_1+{ }^n C_2=36
\Rightarrow \quad{ }^{n+1} C_2=\frac{9 \times 8}{2}={ }^9 C_2

\therefore \quad n=8

\text { From Eq. (i), } \frac{7}{2} \cdot\left(2^x\right)^{-3}=7

or

\begin{aligned} \left(2^x\right)^{-3} & =2 \\ -3 x & =1 \\ x & =-\frac{1}{3} \end{aligned}

 

Posted by

shivangi.shekhar

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