College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.6 - The Binomial Theorem - 8.6 Exercises: 15

Answer

$\displaystyle ( \frac{1}{x}-\sqrt{x})^{5}=\frac{1}{x^{5}}-\frac{5}{x^{7/2}}+\frac{10}{x^{2}}-\frac{10}{x^{1/2}}+5x-x^{5/2}$ Or with negative exponents: $\displaystyle ( \frac{1}{x}-\sqrt{x})^{5}=x^{-5}-5x^{-7/2}+10x^{-2}-10x^{-1/2}+5x-x^{5/2}$

Work Step by Step

We use the values from the 5th row of Pascal's Triangle to find the coefficients: $\displaystyle ( \frac{1}{x}-\sqrt{x})^{5}=(\frac{1}{x})^{5}-5(\frac{1}{x})^{4}\sqrt{x}+10(\frac{1}{x})^{3}x-10(\frac{1}{x})^{2}x\sqrt{x}+5(\frac{1}{x})^1x^{2}-x^{2}\sqrt{x} =\frac{1}{x^{5}}-\frac{5}{x^{7/2}}+\frac{10}{x^{2}}-\frac{10}{x^{1/2}}+5x-x^{5/2}$ Or with negative exponents: $\displaystyle=x^{-5}-5x^{-7/2}+10x^{-2}-10x^{-1/2}+5x-x^{5/2}$
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