Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 12 - Section 12.6 - The Binomial Theorem - 12.6 Exercises: 12

Answer

$99+70\sqrt{2}$

Work Step by Step

$.................1$ $..............1....2...1$ $............1...3.....3...1.$ $..........1...4....6....4.....1$ $........1...5....10...10....5.....1$ $......1..6...15....20....15.....6.....1$ $(1+\sqrt{2})^{6}=$ $=1\cdot1^{6}(\sqrt{2})^{0}+$ $+6\cdot 1^{5}\cdot(\sqrt{2})^{1}+$ $+15\cdot 1^{4}\cdot(\sqrt{2})^{2}+$ $+20\cdot 1^{3}\cdot(\sqrt{2})^{3}+$ $+15\cdot 1^{2}\cdot(\sqrt{2})^{4}+$ $+6\cdot 1\cdot(\sqrt{2})^{5}+$ $+1\cdot(\sqrt{2})^{6}$ $=1^{6}+6\cdot 1^{5}\cdot\sqrt{2}+15\cdot 1^{4}\cdot 2+20\cdot 1^{3}\cdot 2\sqrt{2}+15\cdot 1^{2}\cdot 4+6\cdot 1\cdot 4\sqrt{2}+2^{3}$ $=1+6\sqrt{2}+30+40\sqrt{2}+60+24\sqrt{2}+8$ $=99+70\sqrt{2}$
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