Question

In the expansion of $(3+kx{)}^9$ the Coefficients of $x^2$ and $x^3$ are equal. Find k.

• A. $\frac{3}{7}$
• B. $\frac{7}{3}$
• C. $\frac{9}{7}$
• D. $\frac{7}{9}$

Answer 1

The correct option is C $\frac{9}{7}$

$Coefficientof{x}^2{=}^9{C}_2{3}^{9-2}(k{)}^2$

$andcoefficientof{x}^3{=}^9{C}_3{3}^{9-3}(k{)}^3$

$or{}^9{C}_2\times 3^7{k}^2{=}^9{C}_3\times 3^6{k}^3$

$\left(\frac{9!}{7!\times 2!}\right)\times 3k^2=\left(\frac{9!}{6!\times 3!}\right)\times k^3\left(\frac{1}{7}\right)\times 3=\left(\frac{1}{3}\right)\times k$

$\therefore k=\left(\frac{9}{7}\right)$