dk 5dk-16dk - 2 for every integer k 2 2. = By definition of do, d₁, d₂, ..., for each integer for each integer k with k ≥ 2, in terms of k, dk = (*) dk-1 = and dk -2 = It follows that for each integer k ≥ 2, in terms of k, 5dk-16dk - 2 = 5 = = = = 3k = dk (**) 2k (***). 3k-1- 3k-1 3k-1- k- 2k-1-2-3-3 2k-1-2.3* 2k-1 by substitution from ? by substitution from (**) and ? +2.3.2 +3.2 k- k- V
dk 5dk-16dk - 2 for every integer k 2 2. = By definition of do, d₁, d₂, ..., for each integer for each integer k with k ≥ 2, in terms of k, dk = (*) dk-1 = and dk -2 = It follows that for each integer k ≥ 2, in terms of k, 5dk-16dk - 2 = 5 = = = = 3k = dk (**) 2k (***). 3k-1- 3k-1 3k-1- k- 2k-1-2-3-3 2k-1-2.3* 2k-1 by substitution from ? by substitution from (**) and ? +2.3.2 +3.2 k- k- V
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:Let do, d₁, d₂, ... be a sequence defined by the formula dn = 3″ – 2n for every integer n ≥ 0. Fill in the blanks to show that do, d₁, d₂, ... satisfies the following recurrence relation.
dk = 5dk - 1
6dk - 2 for every integer k > 2.
By definition of do, d₁, d₂, for each integer k with k ≥ 2, in terms of k,
dk
dk-1 =
and dk - 2
=
5dk - 1
=
6dk - 2
It follows that for each integer k ≥ 2, in terms of k,
)-([
6
= 5
=
=
=
=
-
=
5(C
000
(*)
(**)
(***).
3k - 2k
dk
). 3k - 1
).3K-:
(¯). 3K-
3k - 1
3k-1-
· 2k − 1. ·2·3.3k-[
-
·2k-1-2.3k
.2k-1
by substitution from ?
by substitution from (**) and ?
к
+2.3.2
+ 3.2
k-
k
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