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# binomial theorem

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Find the coefficient of x^2 in the expansion of (3/x^2 - 4x^4)^8.

Aug 15, 2020

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Find the coefficient of x^2 in the expansion of (3/x^2 - 4x^4)^8.

$$T_n=\binom{8}{n}(3x^{-2})^n(-4x^4)^{8-n}\qquad n\in \{0,1,2,3,4,5,6,7,8\}\\ T_n=\binom{8}{n}(3^nx^{-2n})(-4)^{8-n}(x^{32-4n })\\ T_n=\binom{8}{n}(3^n)(-4)^{8-n}(x^{32-4n }x^{-2n})\\ T_n=\binom{8}{n}(3^n)(-4)^{8-n}(x^{32-6n })\\ 32-6n=2\\ 6n=30\\ n=5\\ so\\ \text{You have to sub in n=5 and find the coefficient.}$$

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