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 (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

`(5!)/(2!(5-2)!) 2^5 \left(-\frac92\right)^2 = \boxed{6480}`