If first three terms of $(1 + a x)^{n}$ are $1, 6 x, 16 x^{2}$ respectively, then the value of $(18 + n)^{a}$ is

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Solution

Given $(1 + a x)^{n}$
General term of the expansion is
$T_{r + 1} =^{n} C_{r} (a x)^{r}$
Now, as per given conditions, we get
$T_{1} = 1 \Rightarrow 1 = 1 T_{2} = 6 x \Rightarrow n (a x) = 6 x \Rightarrow a = \frac{6}{n} \dots (1) T_{3} = 16 x^{2} \Rightarrow^{n} C_{2} (a x)^{2} = 16 x^{2} \Rightarrow n (n - 1) a^{2} = 32$
Using equation $(1)$ , we get
$\frac{36 (n - 1)}{n} = 32 \Rightarrow 36 n - 36 = 32 n \Rightarrow n = 9 \Rightarrow a = \frac{2}{3} ∴ (18 + n)^{a} = (27)^{2 / 3} = 9$

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