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Sidana [21]
3 years ago
6

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

Mathematics
1 answer:
il63 [147K]3 years ago
8 0

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 <em>x</em> ³ terms occurs for 5 - <em>k</em> = 3, or <em>k</em> = 2, and its coefficient would be

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

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