Question

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

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

Answer 1

Answer: (C)

$2/3$ and $9$

$(1+ax{)}^n={}^n{C}_{o}1+{}^n{C}_{1}ax+{}^n{C}_2(ax{)}^2$

$\therefore {}^n{C}_1\left(ax\right)=6x$

$\therefore {}^n{C}_2(ax{)}^2=16x^2$

$\therefore \frac{n!}{(n-1)!}\times a=6$

$\therefore xa=6$

$\therefore \frac{n!a^2}{(n-2)!2!}=16$

$\therefore \left(n\right)(n-1)a^2=32$

$\therefore \frac{1}{(n-1)\left(a\right)}=\frac{3}{16}$

$\therefore \frac{1}{na-a}=\frac{3}{16}$

$\therefore 6-a=\frac{16}{3}$

$\therefore a=\frac{6-16}{3}=\frac{2}{3}$

$\therefore x=\frac{6}{2}\times 3=9$

$\therefore x=9,a=\frac{2}{3}$.