In the expansion of ${(2^{x} + \frac{1}{4^{x}})}^{n}$ the sum of the binomial coefficients in the first and the second term is equal to $36$ , and the second term of the expansion is $7$ times as large as the first. Find $x$ .

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Solution

Given
${(2^{x} + \frac{1}{4^{x}})}^{n}$
General term of given expansion
$T_{r + 1} = \sum_{r = 0}^{n}^{n} C_{r} (2^{x})^{n - r} {(\frac{1}{4^{x}})}^{r}$
When $r = 0$ we get first term
$T_{1} =^{n} C_{0} (2^{x})^{n}$
$T_{1} =^{n} C_{0} (2)^{n x}$
When $r = 1$ we get second term
$T_{2} =^{n} C_{1} (2^{x})^{n - 1} (\frac{1}{4^{x}})$
$T_{2} =^{n} C_{1} (2)^{n x - x - 2 x}$
$T_{2} =^{n} C_{1} (2)^{n x - 3 x}$
As given
Sum of coefficient of both term is 36
$^{n} C_{0} +^{n} C_{1} = 36$
$1 + n = 36 \Rightarrow n = 35$
And
$7 \times T_{1} = T_{2}$
$7 \times^{35} C_{0} (2)^{35 x} =^{35} C_{1} (2)^{35 x - 3 x}$
$7 = 35 (2)^{- 3 x}$
$5 = (2)^{3 x}$
Taking log both sides
$log 5 = 3 x log 2$
$log 3 = 3 x \Rightarrow x = 0.15$

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