Question

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

• A. $2,3$
• B. $3,2$
• C. $3/2,4$
• D. $1,6$
• E. $3/2,6$

Answer 1

The correct option is C

$3/2,4$

Explanation for the correct option:

Step 1. Given that, ${(1+ax)}^n=1+6x+(27/2)x^2+a^n{x}^n$ $...\left(i\right)$

The expansion of ${\left(1+ax\right)}^n$ is,

${\left(1+ax\right)}^{\mathbf{n}}\mathbf{=}\mathbf{1}\mathbf{+}\mathit{n}\mathit{a}\mathit{x}\mathbf{+}\frac{\mathbf{n}\left(n-1\right)}{\mathbf{2}\mathbf{!}}{\left(ax\right)}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\left(ii\right)$

Step 2. On comparing the coefficient of like powers of $x$ in Eqs. $\left(i\right)and\left(ii\right).$

$na=6...\left(iii\right)$

$\Rightarrow$ $\frac{27}{2}=\frac{n\left(n-1\right)}{2}.a^2$

$\Rightarrow$ $27=\frac{2n\left(n-1\right)}{2}.a^2$

$\Rightarrow$ $27=\left(n-1\right)\left(na\right).a$

$\Rightarrow$ $27=\left(n-1\right)a6\left[fromEq.\left(iii\right)\right]$

$\Rightarrow$$\left(n-1\right)a=\frac{9}{2}...\left(iv\right)$

From equation (iii) and (iv)

$\frac{\left(n-1\right)6}{n}=\frac{9}{2}$

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

$\Rightarrow$ $4\left(n-1\right)=3n$

$\Rightarrow$$4n-4-3n=0$

$\Rightarrow$ $n-4=0$

$\Rightarrow$ $n=4$

From Equation $\left(iii\right)$, we get

$a=\frac{6}{4}$

$\Rightarrow$$\mathit{a}\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}}\mathbf{}$

$\therefore \mathbf{(}\mathbf{a}\mathbf{,}\mathbf{n}\mathbf{)}\mathbf{=}\mathbf{(}\frac{\mathbf{3}}{\mathbf{2}}\mathbf{,}\mathbf{4}\mathbf{)}$

Hence, option(C) is correct.