Question

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

• A. $2,9$
• B. $3,2$
• C. $2/3,9$
• D. $3/2,6$

Answer 1

Answer: (C)

In binomial expansion of $(1+ax{)}^n$
$T_2={}^n{C}_1{\left(ax\right)}^1$
$\Rightarrow nax=6x$
$\Rightarrow na=6\dots \left(1\right)$
$\Rightarrow n^2{a}^2=36$ (after squaring) $\dots \left(2\right)$
Also, $\frac{n\left(n-1\right)}{2}{a}^2{x}^2=16x^2$
$\Rightarrow n\left(n-1\right)a^2=32\dots \left(3\right)$
Dividing $\left(3\right)$ by $\left(2\right)$ gives
$\frac{n\left(n-1\right)}{n^2}=\frac{32}{36}$
$\Rightarrow \frac{n-1}{n}=\frac{8}{9}$
$\Rightarrow n=9$
$\therefore$ From $\left(1\right)$, we have $9a=6$
$\Rightarrow a=\frac{2}{3}$