Show me the steps of determine blue and information is here step by step 0 and that our assumptionholds for n – 2. That is;kT (foisap + fei-3ah ( feg + fei-ik )Soi+3p + foi+2h) (Toi-19 + foi-2k...


Show me the steps of determine blue and information is here step by step


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

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
Jun 05, 2022
SOLUTION.PDF

Get Answer To This Question

Related Questions & Answers

More Questions »

Submit New Assignment

Copy and Paste Your Assignment Here