Consider the nonhomogeneous linear recurrence relation an=2an-1+27. Click and drag the given steps to their corresponding step names to show that an = n2" is a solution of the given recurrence relation in the correct order. Step 1 Step 2 Step 3 The right-hand side of the given recurrence relation can be written as 2(n-1)2-1+2 If an = n2" for all n, then an-1=2(n-1)2-1). If an = n2" for all n, then an-1=(n-1)20-1), The right-hand side of the given recurrence relation can be written as 2(n-1)2-1)+26-1) Substituting both equations into the recurrence, we obtain (1 1)2-1+2= 12-1 Substituting both equations into the recurrence, we obtain (- 1)2"+2" 2".
Consider the nonhomogeneous linear recurrence relation an=2an-1+27. Click and drag the given steps to their corresponding step names to show that an = n2" is a solution of the given recurrence relation in the correct order. Step 1 Step 2 Step 3 The right-hand side of the given recurrence relation can be written as 2(n-1)2-1+2 If an = n2" for all n, then an-1=2(n-1)2-1). If an = n2" for all n, then an-1=(n-1)20-1), The right-hand side of the given recurrence relation can be written as 2(n-1)2-1)+26-1) Substituting both equations into the recurrence, we obtain (1 1)2-1+2= 12-1 Substituting both equations into the recurrence, we obtain (- 1)2"+2" 2".
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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Transcribed Image Text:Consider the nonhomogeneous linear recurrence relation an=2an-1 + 27.
Click and drag the given steps to their corresponding step names to show that an = n2" is a solution of the given recurrence relation in
the correct order.
Step 1
Step 2
Step 3
The right-hand side of the given recurrence relation can be
written as 2(n-1)2(-1) + 2".
If an = n2n for all n, then an-1= 2(n − 1)2(n − 1).
If an = n2n for all n, then an-1= (n − 1)2(n − 1).
The right-hand side of the given recurrence relation can be
written as 2(n-1)2(-1)+2(-1).
Substituting both equations into the recurrence, we obtain (n-
1)2"-1+2= n2"-1
Substituting both equations into the recurrence, we obtain (n-
1)2"+2" = n2".
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