f(n) = 2f(n – 1) +n² for n > 1; f(0) = 1. %3D -
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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please solve all parts or atleast (d) , (e) and (f)
![A.20. Solve the following recurrence relations.
(a) f(n) = f(n – 1) + n? for n > 1; f(0) = 0.
(b) f(n) = 2f(n – 1) +n for n > 1; f(0) = 1.
(c) f(n) = 3f(n – 1) + 2" for n > 1; f(0) = 3.
(d) f(n) = 2f(n – 1) + n2 for n> 1; f(0) = 1.
(e) f(n) = 2f(n – 1) + n + 4 forn> 1; f(0) = 4.
(f) f(n) = -2f(n – 1) + 2" – n² for n > 1; f(0) = 1.
(g) f(n) = nf(n – 1) +1 for n > 1; f(0) = 1.
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Transcribed Image Text:A.20. Solve the following recurrence relations.
(a) f(n) = f(n – 1) + n? for n > 1; f(0) = 0.
(b) f(n) = 2f(n – 1) +n for n > 1; f(0) = 1.
(c) f(n) = 3f(n – 1) + 2" for n > 1; f(0) = 3.
(d) f(n) = 2f(n – 1) + n2 for n> 1; f(0) = 1.
(e) f(n) = 2f(n – 1) + n + 4 forn> 1; f(0) = 4.
(f) f(n) = -2f(n – 1) + 2" – n² for n > 1; f(0) = 1.
(g) f(n) = nf(n – 1) +1 for n > 1; f(0) = 1.
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