If the coefficients of x² and x³ in the expansion of (3 + ax)⁹ are the same, then the value of a is
(a) $- \frac{7}{9}$

(b) $- \frac{9}{7}$

(c) $\frac{7}{9}$

(d) $\frac{9}{7}$

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Solution

(d) $\frac{9}{7}$
Coefficients of x² = Coefficients of x³

${}^{9}C_{2} \times 3^{9 - 2} a^{2} = {}^{9}C_{3} \times 3^{9 - 3} a^{3} \Rightarrow a = \frac{{}^{9}C_{2}}{{}^{9}C_{3}} \times 3 = \frac{9! \times 3! \times 6! \times 3}{2! \times 7! \times 9!} = \frac{9}{7}$

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