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X6n-2 = r X6n-2 = r n-2 2p+h p+h (2P II i=0 h n-2 2p+h II p+h i=0 (fei+sp + foi+7h) (fo1+39 + foi+2k) f6i+49 +f6i+3k foi+7P + foi+6h, h 2p+h p+h n-2 - ( ² P + h) II ( i=0 (2p+h) h+P n-2 (1 Jei+7p+for+h) (f6i+49+J6i+3k\ f6i+39+f6i+2k n-2 II i=0 (fai+sp + foi+7h) (fot+39 + foi+2k) f6i+49 +f6i+3k\ f6i+7P + foi+ch feng+fen-1kh fon+19+fenk P f6i+8p+f6i+7h f6i+49+f6i+3k f6i+7p+f6i+6h 1) ( f6i+39+f6i+2k h1+ feng+fen-1k fon+19+fenk (f6i+8P + f6i+7h (f6i+49 +f6i+3k\

X6n-2 = r X6n-2 = r n-2 2p+h p+h (2P II i=0 h n-2 2p+h II p+h i=0 (fei+sp + foi+7h) (fo1+39 + foi+2k) f6i+49 +f6i+3k foi+7P + foi+6h, h 2p+h p+h n-2 - ( ² P + h) II ( i=0 (2p+h) h+P n-2 (1 Jei+7p+for+h) (f6i+49+J6i+3k\ f6i+39+f6i+2k n-2 II i=0 (fai+sp + foi+7h) (fot+39 + foi+2k) f6i+49 +f6i+3k\ f6i+7P + foi+ch feng+fen-1kh fon+19+fenk P f6i+8p+f6i+7h f6i+49+f6i+3k f6i+7p+f6i+6h 1) ( f6i+39+f6i+2k h1+ feng+fen-1k fon+19+fenk (f6i+8P + f6i+7h (f6i+49 +f6i+3k\

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.5: The Binomial Theorem
Problem 16E
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Show me the steps of determine blue and information is here step by step. It complete

Brn-1n-2 YIn-1 + 8xn-4' = 0, 1, .., In+1 = aIn-2+ (1) The following special case of Eq.(1) has been studied Xn-1In-2 In+1 = In-2 + (8) In-1 + In-4' where the initial conditions r-4, x-3, x-2, x-1,and ro are arbitrary non zero real numbers. Theorem 4. Let {In}-4 be a solution of Eq.(8). Then for n = 0,1, 2, .. п-1 feip + fei-ih ( fei+29 + foi+1k hII fo X6n-4 i-1p+ foi-2h fei+19 + feik i=0 п-1 fei+4p+ fei+3h feig + foi-ik foi+3P + fei+2h) ( Fei-19 + fei-2k) n-1 ( fei+2P + Joi+" a39 + foi+2k , X6n-3 %3D i=0 fei+49+ fei+3k X6n-2 fei+1P + feih i=0 n-1 foi+24+ fei+1k foi+19 + feik foi+6p+ fei+5h PII foi+5p + fei+ah X6n-1 %3D i=0 n-1 П (fei+4p+ foi+3h\ ( foi+69 + föi+5k foi+3p + fei+2h) X6n %3D foi+59 + foi+ak, i=0 п-1 ( föi+8P+ fei+7h П foi+7P + foi+sh foi+49 + f6i+3k fei+39 + f6i+2k) 2р + h T6n+1 %3D p+h i=0 where x-4 = h, x-3 = k, x-2 = r, x-1 = p, xo = q, {fm}m=-1 Proof: For n = 0 the result holds. Now suppose that n > 0 and that our assumption holds for n – 2. That is; 100 {1,0, 1, 1, 2, 3, 5, 8, ...}). п-2 foi+4p + fei+3h fei+3p + fei+2h) ( foi-19 + fei-2k ) foig + föi-ik X6n-9 %3D i=0 п-2 fei+2P + fei+1h( fei+49 + foi+3k I. fei+1p + feih ) X6n-8 %3D fei+39 + fei+2k, i=0 п-2 П ( foi+6P + fei+sh\ ( foi+29 + foi+ik fsi+sP+ fei+ah, X6n-7 %3D fei+14 + foik i=0 п-2 foi+4P + foi+3h föi+3P + fei+2h ) \foi+59 + foi+ak ) foi+69 + foi+sk` qII X6n-6 %3D i=0 n-2 föi+8P + foi+7h foi+7P + fei+sh föi+49 + fei+3k ) Foi+:39 + foi+2k, 2p + h T6n-5 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:Brn-1n-2 YIn-1 + 8xn-4' = 0, 1, .., In+1 = aIn-2+ (1) The following special case of Eq.(1) has been studied Xn-1In-2 In+1 = In-2 + (8) In-1 + In-4' where the initial conditions r-4, x-3, x-2, x-1,and ro are arbitrary non zero real numbers. Theorem 4. Let {In}-4 be a solution of Eq.(8). Then for n = 0,1, 2, .. п-1 feip + fei-ih ( fei+29 + foi+1k hII fo X6n-4 i-1p+ foi-2h fei+19 + feik i=0 п-1 fei+4p+ fei+3h feig + foi-ik foi+3P + fei+2h) ( Fei-19 + fei-2k) n-1 ( fei+2P + Joi+" a39 + foi+2k , X6n-3 %3D i=0 fei+49+ fei+3k X6n-2 fei+1P + feih i=0 n-1 foi+24+ fei+1k foi+19 + feik foi+6p+ fei+5h PII foi+5p + fei+ah X6n-1 %3D i=0 n-1 П (fei+4p+ foi+3h\ ( foi+69 + föi+5k foi+3p + fei+2h) X6n %3D foi+59 + foi+ak, i=0 п-1 ( föi+8P+ fei+7h П foi+7P + foi+sh foi+49 + f6i+3k fei+39 + f6i+2k) 2р + h T6n+1 %3D p+h i=0 where x-4 = h, x-3 = k, x-2 = r, x-1 = p, xo = q, {fm}m=-1 Proof: For n = 0 the result holds. Now suppose that n > 0 and that our assumption holds for n – 2. That is; 100 {1,0, 1, 1, 2, 3, 5, 8, ...}). п-2 foi+4p + fei+3h fei+3p + fei+2h) ( foi-19 + fei-2k ) foig + föi-ik X6n-9 %3D i=0 п-2 fei+2P + fei+1h( fei+49 + foi+3k I. fei+1p + feih ) X6n-8 %3D fei+39 + fei+2k, i=0 п-2 П ( foi+6P + fei+sh\ ( foi+29 + foi+ik fsi+sP+ fei+ah, X6n-7 %3D fei+14 + foik i=0 п-2 foi+4P + foi+3h föi+3P + fei+2h ) \foi+59 + foi+ak ) foi+69 + foi+sk` qII X6n-6 %3D i=0 n-2 föi+8P + foi+7h foi+7P + fei+sh föi+49 + fei+3k ) Foi+:39 + foi+2k, 2p + h T6n-5 p+h i=0 Now, it follows from Eq.(8) that X6n-6X6n-7 X6n-4 = x6n-7 + X6n-6 + X6n-9 11
п-2 2р + h П föi+49 + f6i+3k` fsi+39 + föi+2k , foi+8P + fei+7h` ) X6n-2 p+h fei+7P + fei+6h, i=0 п-2 -() II 2p+h f6i+8p+f6i+7h fei+7p+f6i+6h i=0 föi+49+f6i+3k f6i+39+f6i+2k h r p+h + fönq+ fên-1k h fon+19+fönk h+ P п-2 foi+8P + fói+7h П fei+7P + fei+6h, ( föi+49 + f6i+3k foi+39 + fei+2k, 2р + h X6n-2 p+h i=0 п-2 2p+h p+h П f6i+8p+f6i+7h fei+7p+f6i+6h f6i+49+ƒ6i+3k fei+39+f6i+2k h r i=0 fönq+ fén-1k fön+19+fénk h|1+ п-2 foi+8P + foi+7h` П foi+7P + fói+6h , föi+49 + f6i+3k` föi+39 + f6i+2k , 2р + h X6n-2 p+h i=0 п-2 2p+h p+h П f6i+8p+f6i+7h f6i+7p+ f6i+6h foi+49+f6i+3k fei+39+f6i+2k i=0 + fon+19+f6nk+fënq+fên–1k fon+19+ fönk п-2 foitsP + fói+7h foi+7P + f6i+6h foi+49 + f6i+3k fei+39 + fei+2k, 2р + h p+h i=0 n-2 2p+h p+h П f6i+8p+f6i+7h f6i+7p+f6i+6h fei+49+f6i+3k foi+39+f6i+2k / i=0 fön+29+fön+1k fön+19+f6nk п-2 2р + h П foi+sP + föi+7h foi+rP + fei+sh ) Foi+39 + foi+2k , (foi+49 + fói+3k fön+19 + fönk 1+ p+h fon+29 + fön+1k ] i=0 п-2 2р + h П foi+8P + f6i+7h` foi+7P+ foi+ch ) \ foi+39 + föi+2k / (foi+49 + fói+3k \ [ fön+29 + fön+1k + fon+19 + fönk = r p+h fön+29 + fön+1k i=0 п-2 föi+8P + f6i+7h` П Jöi+7P + fei+6h ) Cein2g i fn) Tôn+39 + fön+2k] (2р + h foi+49 + föi+3k` p+h [ fön+29 + fön+1k ] i=0 16
Transcribed Image Text:п-2 2р + h П föi+49 + f6i+3k` fsi+39 + föi+2k , foi+8P + fei+7h` ) X6n-2 p+h fei+7P + fei+6h, i=0 п-2 -() II 2p+h f6i+8p+f6i+7h fei+7p+f6i+6h i=0 föi+49+f6i+3k f6i+39+f6i+2k h r p+h + fönq+ fên-1k h fon+19+fönk h+ P п-2 foi+8P + fói+7h П fei+7P + fei+6h, ( föi+49 + f6i+3k foi+39 + fei+2k, 2р + h X6n-2 p+h i=0 п-2 2p+h p+h П f6i+8p+f6i+7h fei+7p+f6i+6h f6i+49+ƒ6i+3k fei+39+f6i+2k h r i=0 fönq+ fén-1k fön+19+fénk h|1+ п-2 foi+8P + foi+7h` П foi+7P + fói+6h , föi+49 + f6i+3k` föi+39 + f6i+2k , 2р + h X6n-2 p+h i=0 п-2 2p+h p+h П f6i+8p+f6i+7h f6i+7p+ f6i+6h foi+49+f6i+3k fei+39+f6i+2k i=0 + fon+19+f6nk+fënq+fên–1k fon+19+ fönk п-2 foitsP + fói+7h foi+7P + f6i+6h foi+49 + f6i+3k fei+39 + fei+2k, 2р + h p+h i=0 n-2 2p+h p+h П f6i+8p+f6i+7h f6i+7p+f6i+6h fei+49+f6i+3k foi+39+f6i+2k / i=0 fön+29+fön+1k fön+19+f6nk п-2 2р + h П foi+sP + föi+7h foi+rP + fei+sh ) Foi+39 + foi+2k , (foi+49 + fói+3k fön+19 + fönk 1+ p+h fon+29 + fön+1k ] i=0 п-2 2р + h П foi+8P + f6i+7h` foi+7P+ foi+ch ) \ foi+39 + föi+2k / (foi+49 + fói+3k \ [ fön+29 + fön+1k + fon+19 + fönk = r p+h fön+29 + fön+1k i=0 п-2 föi+8P + f6i+7h` П Jöi+7P + fei+6h ) Cein2g i fn) Tôn+39 + fön+2k] (2р + h foi+49 + föi+3k` p+h [ fön+29 + fön+1k ] i=0 16
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