Question

In the expansion of $(3+ax{)}^9$ coefficients of $x^2$ and $x^3$ are equal then $a=$

• A. $\frac{9}{7}$
• B. $\frac{7}{9}$
• C. $-\frac{9}{7}$
• D. $-\frac{7}{9}$

Answer 1

Answer: (A)

$\frac{9}{7}$

It is known that ${(r+1)}^{th}$ term, $\left(T_{r+1}\right),$ in the binomial expansion of ${(a+b)}^n$ is given by $T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r.$

Assuming that $x^2$ occurs in the ${(r+1)}^{th}$ term in the expansion of ${(3+ax)}^9,$ we obtain

$T_{r+1}{=}^9{C}_r{\left(3\right)}^{9-r}{=}^9{C}_r{\left(3\right)}^{9-r}{a}^r{x}^r$

Comparing the indices of $x$ in $x^2$ and in $T_{r+1},$ we obtain $r=2.$

Thus, the coefficients of $x^2$ is ${}^9{C}_2{\left(3\right)}^{9-2}{a}^2=\frac{9!}{2!7!}{\left(3\right)}^7{a}^2=36{\left(3\right)}^7{a}^2.$

Arruming that $x^3$ occurs in the ${(k+1)}^n$ term in the expansion of ${(3+ax)}^9,$ we obtain

$T_{k+1}{=}^9{C}_k{\left(3\right)}^{9-k}{\left(ax\right)}^k{=}^9{C}_k{\left(3\right)}^{9-k}{a}^k{x}^k$

Comparing the indices of $x$ in $x^3$ and $T_{k+1},$ we obtain $k=3$

Thus, the coefficient of $x^3$ is ${}^9{C}_3{\left(3\right)}^{9-3}{a}^3=\frac{9!}{3!6!}{\left(3\right)}^6{a}^3=84{\left(3\right)}^6{a}^3$

In is given that the coefficient of $x^2$ and $x^3$ are the same,

$84{\left(3\right)}^6{a}^3=36{\left(3\right)}^7{a}^2\Rightarrow 84a=36\times 3\Rightarrow a=\frac{36\times 3}{84}=\frac{104}{84}\Rightarrow a=\frac{9}{7}$

Thus, the required value of $a$ is $\frac{9}{7}$