C X6n-4 = X6n-7 + X6n-66n-7 X6n-6 + X6n-9
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.2: Exponents And Radicals
Problem 26E
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Question
Show me the steps of determine red and information is here step by step It complete
Transcribed Image Text:п-1
foip + foi-1h
hII
foi-1p+ fói-2h
(fei+29 + fei+1k`
foi+19 + feik
X6n-4
i=0
fei+aP + foi+3" ) g+ foi-2k.
foi+3P+ föi+2h ,
n-1
feig + fei-1k
kII
X6n-3
i=0
п-1
foi+2P + foi+1h
föi+1P + foih
( föit49 + fei+3k`
föi+39 + foi+2k,
X6n-2
i=0
( foi+6P+ fói+5h`
Jõi+5P + feraah ) ( 6i+29 + foi+1k`
( foi+4P + föi+3h`
foi+3P + foi+2h,
n-1
X6n-1
foi+19 + foik
i=0
п-1
q II
(föit64 + f6i+5k`
foi+59 + fei+ak,
X6n
i=0
п-1
( foi+sP+ fei+7h
II
foi +7P + fei+6h ) ( foi+39 + fei +2k,
2р + h
föi+49 + föi+3k`
X6n+1
%3D
p+h
i=0
where x-4 = h, x-3 = k, x-2 =r, x-1 = p, xo = q, {fm}m=-1 = {1,0, 1, 1, 2, 3, 5, 8, ...}.
Proof: For n = 0 the result holds. Now suppose that n > 0 and that our assumption
holds for n – 2. That is;
п-2
foi+4P+ f6i+3h
kII
foi+3P + foi+2h
feiq + fei-1k
X6n-9
f6i-19 + fei-2k )
i=0
n-2
( fei+2P + fsi+1h`
föi+1p+ fesh ) \fsi+39 + fei+2k ,
( föi+49 + f6i+3k`
X6n-8
i=0
п-2
foi+6P + foi+5h`
( föi+29 + f6i+1k`
PII
X6n-7
fei+sP+ föi+ah ) Fei+19+ feik
( foi+4P+ foi+3" ) (59+ feitak,
fei+3P + f6i+2h )
i=0
n-2
foi+69 + f6i+5k`
X6n-6
i=0
п-2
( foi+8P + fei+7h
II
föi+7P + f6i+6h
( fei+4q + fei+3k`
fei+39 + fei+2k )
2р + h
X6n-5
= r
p+h
i=0
Now, it follows from Eq.(8) that
X6n-6X6n-7
X6n-4 = X6n–7 +
X6n-6 + x6n-9
11
||
Transcribed Image Text:Bxn-10n-2
YIn-1 + dxn-4
Xn+1 = axn-2 +
n = 0,1, ...,
(1)
The following special case of Eq.(1) has been studied
Xn-1Xn-2
In+1 = Tn-2 +
(8)
In-1+ Xn-4'
where the initial conditions x-4, x-3, x-2, -1,and xo are arbitrary non zero real
numbers.
Theorem 4. Let {Tn}-4 be a solution of Eq.(8). Then for n = 0, 1, 2,..
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