Introductio in analysin infinitorum

180 DE SERIEBUS LIB. I. $\frac{\frac{a(a+\sqrt{aa+4b})+2b}{2\sqrt{aa+4b}}}{1 - \frac{a+\sqrt{aa+4b}}{2}z} +$ $\frac{\frac{a(\sqrt{aa+4b}-a)-2b}{2\sqrt{aa+4b}}}{1 - \frac{a-\sqrt{aa+4b}}{2}z}$, hinc terminus generalis erit $\frac{a(\sqrt{aa+4b}+a)+2b}{2\sqrt{aa+4b}} (\frac{a+\sqrt{aa+4b}}{2})^n z^n + \frac{a(\sqrt{aa+4b}-a)-2b}{2\sqrt{aa+4b}} (\frac{a-\sqrt{aa+4b}}{2})^n z^n$; ex quo omnium Serierum recurrentium, quarum quisque terminus per duos praecedentes determinatur, termini generales expedite definiri poterunt. EXEMPLUM V. Invenire terminum generalem hujus Seriei $1+z+2z^2+2z^3+3z^4+3z^5+4z^6+4z^7+\&c.$, quae oritur ex fractione $\frac{1}{1-z-zz+z^3} = \frac{1}{(1-z [TRUNCATED]