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For Exercises 3 and 4, solve the recurrence relation subject to the initial conditions. 3. A(1) = 7, A(2) = 18, and A(n) = 6A(n − 1) − 8A(n − 2) for n ≥ 3. 4. F(1) = 8, F(2) = 18, and F(n) = 6F(n-1)-5F(n-2) for n ≥ 3. 5. Find a closed-form solution for the Lucas sequence L(1) = 1, L(2) = 3, and L(n) = L(n-1)+L(n−2) for n ≥ 3.

For Exercises 3 and 4, solve the recurrence relation subject to the initial conditions. 3. A(1) = 7, A(2) = 18, and A(n) = 6A(n − 1) − 8A(n − 2) for n ≥ 3. 4. F(1) = 8, F(2) = 18, and F(n) = 6F(n-1)-5F(n-2) for n ≥ 3. 5. Find a closed-form solution for the Lucas sequence L(1) = 1, L(2) = 3, and L(n) = L(n-1)+L(n−2) for n ≥ 3.

Advanced Engineering Mathematics
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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For Exercises 3 and 4, solve the recurrence relation subject to the initial conditions. 3. A(1) = 7, A(2) = 18, and A(n) = 6A(n − 1) − 8A(n − 2) for n ≥ 3. 4. F(1) = 8, F(2) = 18, and F(n) = 6F(n-1)-5F(n-2) for n ≥ 3. 5. Find a closed-form solution for the Lucas sequence L(1) = 1, L(2) = 3, and L(n) = L(n-1)+L(n−2) for n ≥ 3.
Transcribed Image Text:For Exercises 3 and 4, solve the recurrence relation subject to the initial conditions. 3. A(1) = 7, A(2) = 18, and A(n) = 6A(n − 1) − 8A(n − 2) for n ≥ 3. 4. F(1) = 8, F(2) = 18, and F(n) = 6F(n-1)-5F(n-2) for n ≥ 3. 5. Find a closed-form solution for the Lucas sequence L(1) = 1, L(2) = 3, and L(n) = L(n-1)+L(n−2) for n ≥ 3.
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