Answer to Question #227557 in Algebra for Ann

Question #227557

Find the fourth term in the expansion x^2 -2y^1/2


1
Expert's answer
2021-08-19T17:18:12-0400

(x22y)12=[x2(12yx2)]12=x(12yx2)12Using binomial expansion, we expand the expression above=x[112.2yx2+12(121).(2yx2)22!+12.(121).(122).(2yx2)23!]=x[1yx2y22x4y33x6]=xyxy22x3y33x5+...(x^2-2y)^{\frac{1}{2}}=[x^2(1-\frac{2y}{x^2})]^{\frac{1}{2}} \\=x(1-\frac{2y}{x^2})^{\frac{1}{2}} \\\text{Using binomial expansion, we expand the expression above} \\=x[1-\frac{1}{2}.\frac{2y}{x^2} + \frac{\frac{1}{2}(\frac{1}{2}-1).(\frac{2y}{x^2})^2}{2!}+ \frac{\frac{1}{2}.(\frac{1}{2}-1).(\frac{1}{2}-2).(\frac{2y}{x^2})^2}{3!}] \\=x[1-\frac{y}{x^2}-\frac{y^2}{2x^4}-\frac{y^3}{3x^6}] \\= x-\frac{y}{x}-\frac{y^2}{2x^3}-\frac{y^3}{3x^5}+...


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