+0

# binomial theorem

0
62
2

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

Guest Aug 14, 2017
Sort:

#1
+90192
0

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
+26102
0

Perhaps the expression is meant to be:  [(x^2 - 2y)/x]^8  in which case:

.

Alan  Aug 14, 2017

### 7 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details