Question

If $(1+ax{)}^n=1+9x+27x^2+\dots \dots$ then $(a,n)=$

• A. $(1,9)$
• B. $(2,9)$
• C. $(3,3)$
• D. $(2,3)$

Answer 1

The correct option is C $(3,3)$

We know $T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$
Applying to the above question, we get
$T_2=9x$
Therefore $9x={}^n{C}_{1}ax$
$9=an$ ...(i)
And $T_3=27x^2$
$27x^2={}^n{C}_2{a}^2{x}^2$
$\frac{n(n-1)}{2}{a}^2=27$
$n(n-1)a^2=54$ ...(ii)
$\therefore 9(an-a)=54$ ...from (i)
$an-a=6$
$9-6=a$
$\therefore a=3$
Substituting in (i), we get
Therefore $n=3$ ...(from (i))