Question

If $(1+ax{)}^n=1+8x+24x^2+...;$ then $\frac{a-n}{a+n}$ is equal to

• A. $3$
• B. $\frac{-1}{3}$
• C. $-3$
• D. $\frac{1}{3}$

Answer 1

Answer: (B)

$\frac{-1}{3}$

$(1+ax{)}^n=1+n\left(ax\right){+}^n{C}_2{\left(ax\right)}^2{+}^n{C}_3{\left(ax\right)}^3....$

According to question

$(1+ax{)}^n=1+8x+24x^2+...$

$\Rightarrow an=8$

$a^2=\frac{\left(n\right)(n-1)}{2}=24$

$\Rightarrow a^2\left(n\right)(n-1)=48$

$\Rightarrow a(n-1)=6$

$\Rightarrow \frac{n}{n-1}=\frac{8}{6}=\frac{4}{3}$

$\Rightarrow n=4,\Rightarrow a=2$

$\therefore \frac{a-n}{a+n}=\frac{2-4}{2+4}=\frac{-2}{6}=\frac{-1}{3}$