Therefore n-1 ( fei+29 + fei+1k\ feip + fei-1h fei-1P+ foi-2h ) foi+19 + foik Iên-4 = hI i-0 Also, from Eq.(8), we see that I6n-526n-6 I6n-3 = X6n-6 + X6n-5 + *6n-8 n-2 (fsi+aP + foi+3h\ ( foi+69 +...

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Extracted text: Therefore n-1 ( fei+29 + fei+1k\ feip + fei-1h fei-1P+ foi-2h ) foi+19 + foik Iên-4 = hI i-0 Also, from Eq.(8), we see that I6n-526n-6 I6n-3 = X6n-6 + X6n-5 + *6n-8 n-2 (fsi+aP + foi+3h\ ( foi+69 + fei+sk) föi+3P + fei+2h ) (foi+59 + feitak ) i=0 n-2 n-2 im0 n-2 n-2 Teit39+lei+2k 13 n-2 (fei+4P + foi+3h\ ( foi+69 + foi+sk) П Soi+3P + foi+2h/ %3D Soi+54 + fos+ak , im0 n-2 fes-4p+ fei-sh Joi+ap+Sai+ah i=0 p+h Jest39+Se+ak n-2 n-2 n-2 foi+7p+S6i+ch II (i+sp+Sei+7h 2p+h pth II foitsp+fei+7h S6i+7p+Sei+sh Sei+ip+Seih i=0 i=0 n-2 2p+h pth q II n-2 Jei+s9+Sei+sk föi+4P + fei+3h\ ( foi+69 + fei+5k\ Sei+3P + fei+2h ) \Sei+59 + foi+ak / i=0 «II n-2 feiszp+ feiteh Sei+ip+Seih i=0 2p+h Jeitsp+S6i17h ) n-2 2p+h pth Sei+sp+ fei+zh i-0 Ssi469+S6i+5k Seitng+ Sastak n-2 T fei+4P + fei+3h (fei+69 + fei+sk) fei+3p - "oi+2h, foi+59 + foi+ak, 2p+h + pth 2p+h p+h fón -5p+Són -6h Sen-4p+ fen-sh i-0 n-2 Jei43p+Sei+2h i-0 J6i+59+S6i+ak n-2 ( fei+aP+ fsi+3h \ fsi+3P + fei+2h )fsi+59 + feitak, fei+69 + fei+sk\ 1+ fon-sp+S6n-sh Jén-4p+Sên-sh i-0 n-2 n-2 Jeitsa+ Seitak fei+4P + fei+3h ( fei+69 + foi+sk = 911(Fei+-3p + fei+2h ) \ foi+s9 + foi+ak, + fen-4p+ fen-sh+/en-sp+fon-sh Sen-4P+Sen-sh i=0 n-2 feit4p+fei+3h feisog+fei+sk n-2 П (foi+4P + fei+3h\ ( fei+69 + fei+sk) Soi+3P + foi+2h) ( fei+s9 + foi+ak, i=0 + Sen-ap+ fen-4h Jon-4p+Sen-sh i=0 n-2 foi+ap + fei+3h\ ( foi+69 + foi-sk foi+3p + foi+2h ) (foi+59 + fei-ak, q II fon-4P+ fon-sh 1+ \ fon-3p + fon-gh )| i=0 14


Extracted text: Brn-1In-2 YIn-1 + &xn-4 In+1 = a1n-2 + n = 0, 1,..., (1) 1 The following special case of Eq.(1) has been studied In-1In-2 Int1 = In-2+ (8) Tn-1 + In-4 where the initial conditions r-4, I-3, T-2, T-1,and xo are arbitrary non zero real numbers. Theorem 4. Let {zn}-4 be a solution of Eq.-(8). Then for n = 0, 1,2, ... fep + fes-ih ( foi+29 + foiik Jei-1p+ fei-zh) fo419 + Sauk) farq + Sai-1k feirap + forah ) To-19 + foi-ak , (Sei+aP + fot+3h) fei-2P + fei+1h (foi+19 + Sei+ak = rT fei+1P+ fesh ) Joi+39 + fei+2k ) (fei+ep + fes+sh\ (foi+29 + fei+ik Soi-sp + feisah) fei+ap + fei+ah\ ( feiseq + fei-sk Sai+ap + fei+2h ) Jai459 + Seisak ) fei+19 + fask 2p +h II ap+ fesch) (Tars9 + Suvzk, foi+sP + fouth ( Seisa9 + Soirak In+1 = p+h where r-4 = h, r-3 = k, z-2 = r, r-1 = P, 2o = q, {fm}-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; feirap + foi+ah Jei+ap + fei42h) J6i-19 + fei-2k fauq + foi-1 k i-0 fai+1p + fesh ) fet439 + feirak) Joi+ep + fei+sh (fei+29 + foi+zk foi+sp + fes+ah) Iom-6 = 9TT(fei-4p + foi+3h (fei+09 + fei-sk + fei+zh) \Torr39 + fei+ak ) * (2p + h ( foi-sp+ fesuzh (foir49 + foi+ak Ji+7p + Soisgh) Tausg + Soi+zk p+h Now, it follows from Eq.(8) that IGn-4 = Iộn-7 + 11 feisep + fo+sh ( foi+29 + fe+ik Sunsp + Seissh ) (Tus1g+ fesk) = p|| •II )( )II( ) ) (foi+op + fou-sh ( fei-29 + fei+ikY 1foi+sp + foi+sh) foi+19 + fosk + A PIT fasop + foish) ( for-29 + Sousik Joi4sp+fostah) n-2 = p Fei+sp + feirah) feir19 + faik pg ( foi+6P + fei+sh) ( fei+29 + fei-ik =PlI(asp+ fash)( Far1q + Sak ) 12 PII ´ fouvap + faunh\ ( fa-29 + feieik) Sos+19 + Souk PIIusp + Jourah) 1+ ( PII( ) ) -PT (furep + fouosh (fau29 + fuosk Sei-sp + foi-gh ) fois19 + fuk) n-2 PII( (unt Jeusp + Sanah) Sesig+ fauk (An ( fesep + fursh ( fev29 + Jairak1+a+ fon-ek - PII Sen-o9 + fon-zk] n-2 Sen-s9 + Son -ak-" Sei+ep + Sei+sh) ( Su-29 + Sai+ik [ Sen-sq + Sen-sk = PII(sp+ fursh)Tuer9+ feck ) [Ton-39 + Sen-sk] Therefore R-1 Also, from Eq.(8), we see that Tusap + Seurzli ) ( Jonnog + Senek ) ( )--II( )( ) 13