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n-1 n-1 k n-2 n-2 - = P k X6n-10 X6n-9 = 4 - n-1 п-1 n-1 h k k n-1 X'6n-8 = r X'6n-7 = P h - . k - 4 n-1 n-1 n-1 h k - X6n-6 X6n-5 = r %3D h-P

n-1 n-1 k n-2 n-2 - = P k X6n-10 X6n-9 = 4 - n-1 п-1 n-1 h k k n-1 X'6n-8 = r X'6n-7 = P h - . k - 4 n-1 n-1 n-1 h k - X6n-6 X6n-5 = r %3D h-P

College Algebra (MindTap Course List)
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter8: Sequences, Series, And Probability
Section8.5: Mathematical Induction
Problem 42E
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Show me the steps of determine yellow and information is here step by step .it complete

Proof: For n = 0 the result holds. Now suppose that n > 0 and that our assumption holds for n – 1. That is; п-1 п-1 k п-2 п-2 G) p-h X6n-10 X6n-9 = q k - q п-1 n-1 п-1 h k п-1 (E) 9 - k X6n-8 h - P X6n-7 = p k - q п-1 п-1 р — h п-1 h X6n-6 X6n-5 = r h — р Now, it follows from Eq.(11) that x6n-6X6n-7 X6n-4 = X6n-7 X6n-6 - X6n-9 п-1 п-1 p-h п-1 п-1 k k-q п-1 k п-1 = p п-1 п-1 п-1 п-2 p-h q p-h п-1 n-1 (1) (E) п-1 k п-1 (1) - 1 п-1 n-1 (1) п-1 k п-1 k q-k k п-1 k п-1 1 q - k p п-1 k п-1 k k - 4, k - . Therefore k n-1 X6n-4 = p k. Also, from Eq.(11), we see that X6n-5X6n-6 X6n-3 = x6n-6 X6n-5 - X6n-8 п-1 п-1 п-1 h q-k p-h п-1 п-1 р — Һ h-p = q n-1 п-1 п-1 q-k - r h-p 22
Transcribed Image Text:Proof: For n = 0 the result holds. Now suppose that n > 0 and that our assumption holds for n – 1. That is; п-1 п-1 k п-2 п-2 G) p-h X6n-10 X6n-9 = q k - q п-1 n-1 п-1 h k п-1 (E) 9 - k X6n-8 h - P X6n-7 = p k - q п-1 п-1 р — h п-1 h X6n-6 X6n-5 = r h — р Now, it follows from Eq.(11) that x6n-6X6n-7 X6n-4 = X6n-7 X6n-6 - X6n-9 п-1 п-1 p-h п-1 п-1 k k-q п-1 k п-1 = p п-1 п-1 п-1 п-2 p-h q p-h п-1 n-1 (1) (E) п-1 k п-1 (1) - 1 п-1 n-1 (1) п-1 k п-1 k q-k k п-1 k п-1 1 q - k p п-1 k п-1 k k - 4, k - . Therefore k n-1 X6n-4 = p k. Also, from Eq.(11), we see that X6n-5X6n-6 X6n-3 = x6n-6 X6n-5 - X6n-8 п-1 п-1 п-1 h q-k p-h п-1 п-1 р — Һ h-p = q n-1 п-1 п-1 q-k - r h-p 22
Bxn-10n-2 Xn+1 = axn-2 + n = 0,1, .., (1) YIn-1 + dxn-4 1. The following special case of Eq.(1) has been obtained is follows Xn-1Xn-2 Xn+1 = Xn-2 - (11) Xn-1 - Xn-4 where the initial conditions x-4, x_3, x-2, a-1, xo are arbitrary non zero real num- bers with x-4 # x-1 and x-3 # xo. Theorem 7. Let {rn}-3 be a solution of Eq.(11). Then every solution of Eq.(11) is unbounded. Moreover {x,}03 takes the form =-3 k n-1 п-1 X6n-4 = p X6n-3 = 4 k - h k X6n-2 X6n-1 = p n+1 (). h X6n X6n+1 = r %3D р h - p
Transcribed Image Text:Bxn-10n-2 Xn+1 = axn-2 + n = 0,1, .., (1) YIn-1 + dxn-4 1. The following special case of Eq.(1) has been obtained is follows Xn-1Xn-2 Xn+1 = Xn-2 - (11) Xn-1 - Xn-4 where the initial conditions x-4, x_3, x-2, a-1, xo are arbitrary non zero real num- bers with x-4 # x-1 and x-3 # xo. Theorem 7. Let {rn}-3 be a solution of Eq.(11). Then every solution of Eq.(11) is unbounded. Moreover {x,}03 takes the form =-3 k n-1 п-1 X6n-4 = p X6n-3 = 4 k - h k X6n-2 X6n-1 = p n+1 (). h X6n X6n+1 = r %3D р h - p
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