Show me the steps of determine yellow and inf is here The main aim of this study is to exhibit some cases on the periodic character ofthe positive solutions of the rational difference equationaSn-q + bSn-r + cSn-sdSnSn+1 = Sn-p(1)+ eSn-r + fSn-s ) '-qwhere a, b, c, d, e, ƒ€ (0, 0). The initial conditions S-p, S-p+1;.-,S-q, S-q+1;..,S-r,S-r+1,...,S-s,...,S_s+1,...,S_1 and So are arbitrary positive real numbers such thatp > q > r > s > 0.е, Case 3. Suppose the positive integer p,q,r are even and the positive integer s isodd. In this caseSn = Sn-p = Sn-q = Sn-r and Sn+1Sn-s.From Equation (1), we have$ = ( a+b)b + có(d+ e) + fø,(а + b)ф + сф(d+ e)o+ f )f = ¢Thus,(d + e)o + fo? = (a + b)y² + cop,(6)%3|and(d+e)ob + fv? = (a + b)o? + cop.(7)By subtracting (6) from (7), we deduce thatf(4? – v?) + (a + b)(? – y²) = 0.Hence, we have(a + b+ f)(@? – u²) = 0.%3DIf ø + 4, then(a + b+ f) == 0.Since a, b,and f are nonzero positive real numbers, thus (a + b+ f) # 0. Thisimplies ø = v. This contradicts the hypothesis o + 4.

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Case 3. Suppose the positive integer p,q,r are even and the positive integer s is odd. In this case Sn = Sn-p = Sn-q = Sn-r and Sn+1 = Sn-s- From Equation (1), we have (a + b) + co (d+ e) + f¢, (a+b)o+cb (d+e)o+ fw, $ = b b = $ (: Thus, (d+ e)ob + fo? = (a + b)? + cop, (6) %3D and (d+ e)ob + fp? = (a + b)o? + conp. (7) By subtracting (6) from (7), we deduce that f(6? – v²) + (a + b)(4² – v²) = 0. %3D Hence, we have (a + b+ f)(6? – v²) = 0. If ø + v, then (a + b+ f) = 0. Since a, b,andf are nonzero positive real numbers, thus (a + b+ f) # 0. This %3D