Unformatted text preview: 3 , the right side is ~. Since 0 < ~,the statement is true. Now assume that k ~ 1 and the statement is true for n = k; that is, assume We must prove that the statement is true for n = k + 1; that is, We have 1 3 + 2 3 + 3 3 + . .. + k 3 = 1 3 + 2 3 + 3 3 + . .. + (k _1)3 + k 3 k4 < '4 + k 3 , using the induction hypothesis, k4 + 4k 3 k4 + 4k 3 + 6k 2 + 4k + 1 (k + 1)4 4 < 4 4 which is the desired result. By the Principle of Mathematical Induction, inequality (*) holds for all n~1. Next we prove that for all n ~ 1. Since 1 > ~, the inequality is true for n = 1. Now assume k ~ 1 and the inequality is true for n = k; that is,...
View
Full Document
 Summer '10
 any
 Binomial Theorem, Graph Theory, Mathematical Induction, Binomial, binomial expansion, 3k, 2k, 3 k

Click to edit the document details