The first 3 terms in the expansion of $(1 + ax)^{n}\ (n
eq 0)$ are $1,\ 6x,$ and $16x^{2}$. Then, the value of $a$ and $n$ are respectively

Question

The first 3 terms in the expansion of $(1 + ax)^{n}\ (n 
eq 0)$ are $1,\ 6x,$ and $16x^{2}$. Then, the value of $a$ and $n$ are respectively

Dr. Abubacker · Accepted Answer

(c)
Given that, $T_{1} = \ ^{n}C_{0} = 1$ …(i)
$T_{2} = \ ^{n}C_{1}ax = 6x$
$\Rightarrow \ \ \frac{n!}{(n - 1!)}a = 6 \Rightarrow na = 6$ …(ii)
and $T_{3} = {\ ^{n}C}_{2}(ax)^{2} = 6x^{2}$
$\Rightarrow \ \frac{n(n - 1)}{2}a^{2} = 16$ ...(iii)
Only option (c) is satisfying Eqs. (ii) and (iii)

Q. The first 3 terms in the expansion of $(1 + ax)^{n}\ (n \neq 0)$ are $1,\ 6x,$ and $16x^{2}$. Then, the value of $a$ and $n$ are respectively

Explanation

Reviewed by Dr. Abubacker

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