By the inductive hypothesis and the definition of divisibility, there exists an integer r such that 7k – 2k = 5r. Then 7k + 1 - 2k + 1 = 7 · 7k – 2· 2k = (5 + 2) · 7k – 2 · 2k. Continue simplifying the right-hand side of the equation, apply the induction hypothesis, and express the result in terms of k and r. 7k + 1- 2k + 1 - 5 . zk + 2r This quantity ic an inf
By the inductive hypothesis and the definition of divisibility, there exists an integer r such that 7k – 2k = 5r. Then 7k + 1 - 2k + 1 = 7 · 7k – 2· 2k = (5 + 2) · 7k – 2 · 2k. Continue simplifying the right-hand side of the equation, apply the induction hypothesis, and express the result in terms of k and r. 7k + 1- 2k + 1 - 5 . zk + 2r This quantity ic an inf
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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How did they get 5(7^k + 2r)

Transcribed Image Text:By the inductive hypothesis and the definition of divisibility, there exists an integer r such that 7k – 2k = 5r. Then
7k + 1 - 2k + 1 = 7 · 7k – 2· 2k = (5 + 2) · 7k – 2 · 2k.
Continue simplifying the right-hand side of the equation, apply the induction hypothesis, and express the result in terms of k and r.
7k + 1- 2k + 1 - 5 .
7k
+ 2r
This quantity ic an into
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