Introductio in analysin infinitorum

RECURRENTIBUS. 193 CAP. XIII. 229. Simili modo, si statuantur termini sequentes $A+Bz+Cz^2+.....+Pz^n+Qz^{n+1}+Rz^{n+2}+.....+$ $X z^{2n} + Y z^{2n+1} + Z z^{2n+2},$ erit $Z = \frac{(\mathfrak{C}A-\alpha B)Q^2+2BQR-ARR}{BB-\alpha AB+\mathfrak{C}AA}, \&, \text{ob } R=\alpha Q - \mathfrak{C}P.$ erit $Z = \frac{-\mathfrak{C}\mathfrak{C}AP^2+2\mathfrak{C}(\alpha A-B)PQ+(\alpha B-(\alpha\alpha-\mathfrak{C})A)Q^2}{BB-\alpha AB+\mathfrak{C}AA}$ At est $Z = \alpha Y - \mathfrak{C}X,$ ergo $Y = \frac{Z+\mathfrak{C}X}{\alpha};$ unde fit $Y = \frac{-\mathfrak{C}BP^2+2\mathfrak{C}APQ+\alpha(B-\alpha A)QQ}{BB-\alpha AB+\mathfrak{C}AA}.$ Sic igitur porro ex $X$ & $Y$ definiri poterunt simili modo coëfficientes potestatum $z^{4n}, \& z^{4n+1};$ hincque ipsarum $z^{8n}, z^{8n+1}, \&$ ita porro. EXEMPLUM. Sit proposita ista Series recurrens $1+3z+4z^2+7z^3+11z^4+18z^5+.....+Pz^n+Qz^{n+1}+\&c.,$ cujus cum quilibet coëfficiens fit summa duorum præcedentium, erit denominator fractionis hanc Seriem producentis $1-z-zz;$ ideoque $\alpha=1; \mathfrak{C}=-1; \& A=1; B=3;$ unde fit $BB-\alpha AB+\mathfrak{C}AA=5,$ ex quo orietur primum $Q = \frac{P+\sqrt{(5PP+20(-1)^n)}}{2}$ $\frac{P+\sqrt{(5PP \pm 20)}}{2},$ ubi signum superius valet, si $n$ sit numerus par, inferius si impar. Sic, si $n=4,$ ob $P=11,$ erit Euleri Introduct. in Anal. infin. parv. Bb $Q=$