Introductio in analysin infinitorum

194 DE SERIEBUS LIB. I. $Q = \frac{11+\sqrt{(5 \cdot 121+20)}}{2} = \frac{11+25}{2} = 18$. Si porro coëfficiens termini $z^{2n}$ fit $X$, erit $X = \frac{4PP+6PQ-QQ}{5}$; ergo $X = \frac{-4 \cdot 121+6 \cdot 198-324}{5} = 76$. Cum autem fit $Q = \frac{P+\sqrt{(5PP \pm 20)}}{2}$ erit $QQ = \frac{3PP \mp 10+P\sqrt{(5PP \pm 20)}}{2}$; ideoque $X = \frac{-PP \mp 2+P\sqrt{(5PP \pm 20)}}{2}$. Ex termino ergo Seriei quocunque $Pz^n$, obtinentur hi $\frac{P+\sqrt{(5PP \pm 20)}}{2} z^{n+1}$, & $\frac{-PP \mp 2+P\sqrt{(5PP \pm 20)}}{2} z^{2n}$. 230. Simili modo in Seriebus recurrentibus, quarum quilibet terminus ex tribus antecedentibus determinatur, quivis terminus ex duobus antecedentibus definiri potest. Sit enim Series hujusmod [TRUNCATED]