In the expansion of ${(2^{x} + \frac{1}{4^{x}})}^{n}$
(i) $\frac{T_{3}}{T - 2} = 7$
(ii) The sum of the coefficient of 2nd and 3rd term is 36 . Find the value of x

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Solution

$\frac{T_{3}}{T_{2}} = 7 \Rightarrow \frac{n - r + 1}{r} \frac{a}{x} = 7$
Put $r = 2, x = 2^{x}, a = \frac{1}{4^{x}} = 2^{- 2 x} ∴ \frac{a}{x} = 2^{- 3 x}$
$∴ \frac{n - 1}{2} (2^{- 3 x}) = 7$ ....(1)
Also $^{n} C_{1} +^{n} C_{2} = 36 \Rightarrow n + \frac{n (n - 1)}{2} = 36$
or $n^{2} + n - 72 = 0$
or $(n + 9) (n - 8) = 0 ∴ n = 8$
Putting the value of n in (1) , we get
$\frac{7}{2} (2^{- 3 x}) = 7 o r 26^{- 3 x} = 2^{1}$
$∴ - 3 x = 1 o r x = \frac{- 1}{3}$

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