In the expansion of (x-(3/x^2)^9 find the following:
a) the term containing x6
b) the constant term
This is my last question, I swear! :P
The general term is $\displaystyle {9 \choose r} x^{9 - r} \left( -\frac{3}{x^2}\right)^r = {9 \choose r} (-3)^r x^{9 - r} x^{-2r} = {9 \choose r} (-3)^r x^{9 - 3r}$.
a) You require the coefficient of the power of x corresponding to 9 - 3r = 6.
b) You require the coefficient of the power of x corresponding to 9 - 3r = 0.