Answer to Question #227557 in Algebra for Ann

Question

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

Answer 1

$(x^2-2y)^{\frac{1}{2}}=[x^2(1-\frac{2y}{x^2})]^{\frac{1}{2}} \\=x(1-\frac{2y}{x^2})^{\frac{1}{2}} \\\text{Using binomial expansion, we expand the expression above} \\=x[1-\frac{1}{2}.\frac{2y}{x^2} + \frac{\frac{1}{2}(\frac{1}{2}-1).(\frac{2y}{x^2})^2}{2!}+ \frac{\frac{1}{2}.(\frac{1}{2}-1).(\frac{1}{2}-2).(\frac{2y}{x^2})^2}{3!}] \\=x[1-\frac{y}{x^2}-\frac{y^2}{2x^4}-\frac{y^3}{3x^6}] \\= x-\frac{y}{x}-\frac{y^2}{2x^3}-\frac{y^3}{3x^5}+...$