Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.5 - The Binomial Theorum - Exercise Set - Page 1093: 74


The provided sentence makes sense.

Work Step by Step

We know that by using Pascal’s Triangle, we can find the coefficient of each term without expansion of the given function. Therefore, every term having the odd power of $\left( -1 \right)$ will be negative and the terms having even powers of $\left( -1 \right)$ will be positive. Therefore, the first, third, fifth and seventh terms will be positive. And the second, fourth and sixth terms will be negative. Rewrite the function ${{\left( x-1 \right)}^{6}}$ as ${{\left( x+\left( -1 \right) \right)}^{6}}$. Then, by applying Pascal’s Triangle: Positive terms: ${{x}^{6}},15{{x}^{4}}{{\left( -1 \right)}^{2}},15{{x}^{2}}{{\left( -1 \right)}^{4}},{{\left( -1 \right)}^{6}}$ After solving the terms: ${{x}^{6}},15{{x}^{4}},15{{x}^{2}},1$ Negative terms: $6{{x}^{5}}{{\left( -1 \right)}^{1}},20{{x}^{3}}{{\left( -1 \right)}^{3}},6{{x}^{1}}{{\left( -1 \right)}^{5}}$ Then, after solving the terms: $-6{{x}^{5}},-20{{x}^{3}},-6{{x}^{1}}$ Thus, the statement makes sense and the positive and negative terms are ${{x}^{6}},15{{x}^{4}},15{{x}^{2}},1,-6{{x}^{5}},-20{{x}^{3}},-6{{x}^{1}}$.
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