Introductio in analysin infinitorum

RECURRENTIBUS. 179 CAP. XIII. Ob denominatorem $=(1-2z)(1-3z)$ resolvitur fractio in has $-\frac{1}{1-2z} + \frac{2}{1-3z}$, ex quibus fit terminus generalis $2 \cdot 3^n z^n - 2^n z^n = (2 \cdot 3^n - 2^n)z^n$. EXEMPLUM III. Invenire terminum generalem Seriei hujus $1 + 3z + 4z^2 + 7z^3 + 11z^4 + 18z^5 + 29z^6 + 47z^7 + \&c.$, quae oritur ex evolutione fractionis $\frac{1+2z}{1-z-zz}$. Ob denominatoris Factores $1 - (\frac{1+\sqrt{5}}{2})z$ & $1 - (\frac{1-\sqrt{5}}{2})z$, per resolutionem prodeunt $\frac{\frac{\sqrt{5}+1}{2}}{1-(\frac{1+\sqrt{5}}{2})z} + \frac{\frac{1-\sqrt{5}}{2}}{1-(\frac{1-\sqrt{5}}{2})z}$, unde erit terminus generalis $= (\frac{1+\sqrt{5}}{2})^{n+1}z^n [TRUNCATED]