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Extracted text: Br,-10n-2 YIn-1 + 8xn-4' In+1 = QI,n-2 + n = 0, 1,..., (1) 1 The following special case of Eq.(1) has been studied In-1Tn-2 In+1 = In-2 + (8) In-1 + In-4 where the initial conditions r-4, x-3, T-2, r-1,and ro are arbitrary non zero real numbers. Theorem 4. Let {rn}-4 be a solution of Eq.(8). Then for n = 0, 1,2, ... Ton-4 = hT foip + foi-1h foi-1p + foi-ah) ( fei+29 + Soi+2k foi+19 + ferk n-1 n-1 ( foisap + foi+ah foi+3P+ fei+2h) ( fei-19 + fei-2k) fe:9 + fei-1k I'ên-3 = kI i=0 T( foi-2p + foi+h) ( feir49 + fei+zk fei+39 + foi+2k) fei+29 + fei+1k). fei+19 + fesk Ten-2 = foi+1P + foih foi+ep + fei+sh Soi+sp + fei+sh , ( foi+ap + foi+ah ( foi+69 + foi+sk). Sei+ap + fei+2h) fei+59 + fei+ak ) (2p + h (fei+sP+ foi+zh (feit49 + fei+3k n-1 p+h) th II(a-p+farch) (Fars9 + Sa+zk ) where r-4 = h, x-3 = k, x-2 = r, -1 = P, xo = q, {fm}- = {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; kT (foisap + fei-3ah ( feg + fei-ik ) Soi+3p + foi+2h) (Toi-19 + foi-2k ) fei+2P + fei+1h ( fei+1p + feih ) \Foi+39 + fois+2k) fei+ep+ fei+sh ( fei+29 + foi+ik Sei+sp + fei+gh Ten-9 = i-0 Sei+49 + fei+3k n-2 foi+19 + Sauk ) (foisap + fo+3gh ( fei+69 + foi-sk) foi+ap + fei+zh ) Jei459 + foitak) I6n-6 = (2p +4) T(foi-ap + fes-zh) (foisaq + foi+sk Jei+7p+ fei+ch) Joi+39 + fei+2k ) I6n-5 = Now, it follows from Eq.(8) that Z6n-4 = T6n-7 + T6n-6 + 2ôn-9 11 foi+6P + fei+sh\ ( Soi+29 + fei+ik ( foi+6P + fei+sh PIIasp+ feirah ) foi+29 + foi+ik n-2 n-2 foi+6P + foi+sh ( foi+29 + fei+ik PIISP+ feigh) Tei+19 + feik ) n-2 paII (st fa) (2 fuk) n-2 Seseg+ fersk Joi-19+fer-ak im0 pa II ( (atl) = PTT (feisap + feirsh) (feis29 + feisak fei+sp + fei+sh, im0 n-2 Josreg+Sesk) (Tei-19+Se-ak i=0 -i )- ( foi+oP+ fsi+sh` Soi+sp + fei+ah) foi+29 + foi+1k fei+19 + feik =PII 12