Question

If ${\left(1+ax\right)}^n=1+6x+\frac{27}{2}{x}^2+\cdots +a^n{x}^n$, then the values of $a$ and $n$ are respectively

• A. $2,3$
• B. $3,2$
• C. $\frac{3}{2},4$
• D. $1,6$
• E. $\frac{3}{2},6$

Answer 1

Answer: (C)

Given,
$(1+ax{)}^n=1+6x+\frac{27}{2}{x}^2+\dots +a^n{x}^n\dots (i)$
The expansion of $(1+ax{)}^n$ is,
$(1+ax{)}^n=1+nax+\frac{n(n-1)}{2!}(ax{)}^2+\dots (ii)$
On comparing the coefficient of like powers of $x$ in Eqs. (i) and (ii),
$na=6\dots (iii)$
$\frac{27}{2}=\frac{n(n-1)}{2}\cdot a^2$
$\Rightarrow 27=(n-1)(na)\cdot a$
$27-(n-1)a6$ [from Eq. (iii)]
$(n-1)a=\frac{9}{2}\dots (iv)$
From Eqs. (iii) and (iv),
$\frac{(n-1)6}{n}=\frac{9}{2}$
$\Rightarrow \frac{n-1}{n}=\frac{3}{4}$
$\Rightarrow 4n-4=3n$
$\Rightarrow n=4$
From Eq. (iii), $a=\frac{6}{4}$
$\Rightarrow a=3/2$