First, you are missing the "+..." indicating that the series is infinite.
But to answer your question, yes, the sum you get, \(\displaystyle a_1x- \frac{9}{6}a_1x^3+ \frac{9^2}{5!}x^5+\cdot\cdot\cdot\) is not equal to sin(3x)- each term is missing a factor of 3. But it is equal to \(\displaystyle \frac{1}{3}sin(3x)\) which is sufficient since you are multiplying by an undetermined constant, anyway.
