College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 9 - Section 9.5 - The Binomial Theorem - 9.5 Assess Your Understanding - Page 676: 17


$y=1; n=5$ $(x+1)^5=\binom{5}{0}x^51^0+\binom{5}{1}x^{4}1^1+\binom{5}{2}x^{3}1^2+\binom{5}{3}x^{2}1^3+\binom{5}{4}x^{1}1^{4}+\binom{5}{5}x^01^5=x^5+5x^4+10x^3+10x^2+5x+1$

Work Step by Step

The binomial theorem expands an algebraic expression in the form of: $(x+y)^n=\binom{n}{0}x^ny^0+\binom{n}{1}x^{n-1}y^1+\binom{n}{2}x^{n-2}y^2+...+\binom{n}{i}x^{n-i}y^i+...+\binom{n}{n-1}x^{1}y^{n-1}+\binom{n}{n}x^0y^n=\sum_{i=0}^{n}\binom{n}{i}x^{n-i}y^i$ Here: $y=1; n=5$ $(x+1)^5=\binom{5}{0}x^51^0+\binom{5}{1}x^{4}1^1+\binom{5}{2}x^{3}1^2+\binom{5}{3}x^{2}1^3+\binom{5}{4}x^{1}1^{4}+\binom{5}{5}x^01^5=x^5+5x^4+10x^3+10x^2+5x+1$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.