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Elements Of Modern Algebra

8th Edition
Gilbert + 2 others
ISBN: 9781285463230

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Section
BuyFindarrow_forward

Elements Of Modern Algebra

8th Edition
Gilbert + 2 others
ISBN: 9781285463230
Textbook Problem

Given the recursively defined sequence a 1 = 0 , a 2 = 30 , and a n = 8 a n 1 15 a n 2 , use complete induction to prove that a n = 5 · 3 n 3 · 5 n for all positive integers n .

To determine

To prove: an=53n35n for all positive integers n by using complete induction where the recursively defined sequence is a1=0,a2=30, and an=8an115an2.

Explanation

Given information:

The recursively defined sequence is a1=0,a2=30, and an=8an115an2.

Formula used:

Strategy: Proof by Complete Induction

1. Basic Step: The statement is proved true for n=a, where a.

2. Induction Hypothesis: For an integer k, the statement is assumed true for all integers m such that am<k.

3. Inductive Step: Under this assumption, the statement is proved to be true for m=k.

Proof:

Consider the statement, “ an=53n35n

For n=3.

a3=8a3115a32

=8a215a1

It is given that a1=0, and a2=30.

a3=8(30)15(0)=240 and 533353=5273125=135375=240.

Therefore, the statement is true for n=3.

Assume that the statement is true for all integers m such that 3m<k, where k is an integer.

am=53m35m

For m=k

To show that ak=53k35k is true.

Now,

ak=8ak115ak2

As k1<k, and k2<k and the statement is true for 3m<k then ak1=53k135k1 and ak2=53k235k2

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