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Solved: Find the solution to each of these recurrence

Discrete Mathematics and Its Applications | 7th Edition | ISBN: 9780073383095 | Authors: Kenneth Rosen ISBN: 9780073383095 37

Solution for problem 17E Chapter 2.4

Discrete Mathematics and Its Applications | 7th Edition

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Discrete Mathematics and Its Applications | 7th Edition | ISBN: 9780073383095 | Authors: Kenneth Rosen

Discrete Mathematics and Its Applications | 7th Edition

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Problem 17E

Find the solution to each of these recurrence relations and initial conditions. Use an iterative approach such as that used in Example 10.

Step-by-Step Solution:

Step 1:

In this problem we have to find the solution for these recurrence relation, where we given the initial condition.

a:  the recurrence relation is given as

 an = 3 an-1 with a0= 2

Now we starting with initial condition a0 = 2 and working upward until we reach an to deduce a closed formula for the sequence. Then

We put n = 0 ,1,2,3,4……...so on , Then

a1= 3a1-1 

a1 = 3 a0 = 3.2

a2 = 3a2-1

a2 = 3a1 = 3.3.2 = 32.2

a3 = 3a3-1

a3 = 3a2 = 3.32.2 = 33.2

………..

………..

an  = 3an-1  = 3n.2

Step 2:

b: the recurrence relation is given as

 an =  an-1 +2  with a0= 3

Now we starting with initial condition a0 = 3 and working upward until we reach an to deduce a closed formula for the sequence. Then

We put n = 0 ,1,2,3,4……...so on , Then

a1= a1-1+2 

a1 =  a0 +2 = 3+2

a2 = a2-1 +2

a2 = a1 +2  = 3+2+2 = 3+2.2

a3 = a3-1+2

a3 = a2 +2  = 3+2.2+2 = 3+3.2

………..

………..

an  = an-1  +2 = 3+ 2n.

Step 3:

c: the recurrence relation is given as

 an =  an-1 +n  with a0= 1

Now we starting with initial condition a0 = 1 and working upward until we reach an to deduce a closed formula for the sequence. Then

We put n = 0 ,1,2,3,4……...so on , Then

a1= a1-1+1

a1 =  a0 +1 = 1+1

a2 = a2-1 +2

a2 = a1 +2  = 1+1+2 = 1+ (1+2)

a3 = a3-1+3

a3 = a2 +3  = 1+1+2+3 = 1+ (1+2+3)

a4 = a3 +4  = 1+1+2+3+4 = 1 + (1+2+3+4 ) = 1+ n(n+1)/2

                                                      (since by A.P. series  1+2+3+.........+n = n(n+1)/2)

 ………..

………..

an  = an-1  + n = 1+ n(n+1)/2.

Step 4:

d: the recurrence relation is given as

 an =  an-1 + 2n +3  with a0= 4

Now we starting with initial condition a0 = 4 and working upward until we reach an to deduce a closed formula for the sequence. Then

We put n = 0 ,1,2,3,4……...so on , Then

a1= a1-1+2.1 +3

a1 =  a0 +2.1+3  = 4+ 2.1 +3

a2 = a2-1 +2.2 +3

a2 = a1 +2.2+3  = 4 +2.1+3+2.2+3 = 4 + 2(1+2) +2.3

a3 = a3-1+2.3+3

a3 = a2 +2.3+3  =4 +2 (1+2+3) +3.3

 ………..

………..

an  = an-1  + 2n +3  = 4 +2( 1+2+...... +n) +n.3= 4+ 2n( n+1)/2 +n.3,      

                                                                          (since by A.P. series  1+2+3+.........+n = n(n+1)/2)                                                  

an  = an-1  + 2n +3 = n2 +4n +4.

Step 5 of 7

Chapter 2.4, Problem 17E is Solved
Step 6 of 7

Textbook: Discrete Mathematics and Its Applications
Edition: 7
Author: Kenneth Rosen
ISBN: 9780073383095

This full solution covers the following key subjects: approach, conditions, Example, Find, initial. This expansive textbook survival guide covers 101 chapters, and 4221 solutions. The answer to “Find the solution to each of these recurrence relations and initial conditions. Use an iterative approach such as that used in Example 10.” is broken down into a number of easy to follow steps, and 23 words. Discrete Mathematics and Its Applications was written by and is associated to the ISBN: 9780073383095. This textbook survival guide was created for the textbook: Discrete Mathematics and Its Applications, edition: 7. Since the solution to 17E from 2.4 chapter was answered, more than 253 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 17E from chapter: 2.4 was answered by , our top Math solution expert on 06/21/17, 07:45AM.

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Solved: Find the solution to each of these recurrence

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