Question

Find the coefficient of the x^3 in the expansion of (2x-9)^5

Answer 1

Use the binomial theorem:

$\displaystyle (2x-9)^5 = \sum_{k=0}^5 \binom5k (2x)^{5-k}(-9)^k = \sum_{k=0}^5 \frac{5!}{k!(5-k)!} 2^5 \left(-\frac92\right)^k x^{5-k}$

The x ³ terms occurs for 5 - k = 3, or k = 2, and its coefficient would be

$\dfrac{5!}{2!(5-2)!} 2^5 \left(-\dfrac92\right)^2 = \boxed{6480}$