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# binomial theorem

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Find the term not involving x for the expansion of (x^2-2y/x)^8

Guest Aug 14, 2017
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Find the term not involving x for the expansion of (x^2-2y/x)^8

$$(x^2-\frac{2y}{x})^8\\ \text{The nth term will be}\\ \binom{8}{n}(\frac{-2y}{x})^n(x^2)^{8-n}\\ =\binom{8}{n}(-2)^ny^n\frac{(x^2)^{8-n}}{x^n}\\ =\binom{8}{n}(-2)^ny^n x^{16-2n-n}\\ =\binom{8}{n}(-2)^ny^n x^{16-3n}\\$$

16-3n=0 has no integer solutions so there is no term that does not involve x.

Melody  Aug 14, 2017
#2
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Perhaps the expression is meant to be:  [(x^2 - 2y)/x]^8  in which case:

.

Alan  Aug 14, 2017