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If in the expansion of \left(2^x+\frac{1}{4^x}\right)^n,  T3/T2 = 7 and the sum of the co-efficient of 2nd  and 3rd term is 36, then the value of x is

Option: 1

 –1/3


Option: 2

–1/2


Option: 3

 1/3


Option: 4

 ½


Answers (1)

best_answer

Given that { }^n C_1+{ }^n C_2=36

\Rightarrow \mathrm{n}=8, \mathrm{n} \neq-9

also \frac{{ }^n C_2\left(2^x\right)^{n-2} \cdot\left(\frac{1}{4 x}\right)^2}{{ }^n C_1\left(2^x\right)^{n-1} \cdot\left(\frac{1}{4 x}\right)^1}=7

\frac{28\left(2^x\right)^6 \cdot\left(\frac{1}{4 x}\right)^2}{8\left(2^x\right)^7 \cdot\left(\frac{1}{4 x}\right)^1}=7 \Rightarrow 8^x=\frac{1}{2} \Rightarrow x=\frac{-1}{3}

Posted by

shivangi.bhatnagar

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