Introductio in analysin infinitorum

PER SERIES INFINITAS. 59 Cum igitur primus terminus fit $= 1$, & antecedentes nulli, CAP. IV. erit $A = \frac{m-1}{1} \alpha$ $B = \frac{(m-2)}{2} \alpha A + \frac{(2m-2)}{2} \mathfrak{b}$ $C = \frac{(m-3)}{3} \alpha B + \frac{(2m-3)}{3} \mathfrak{b} A + \frac{(3m-3)}{3} \gamma$ $D = \frac{(m-4)}{4} \alpha C + \frac{(2m-4)}{4} \mathfrak{b} B + \frac{(3m-4)}{4} \gamma A$ $E = \frac{(m-5)}{5} \alpha D + \frac{(2m-5)}{5} \mathfrak{b} C + \frac{(3m-5)}{5} \gamma B$ &c. 76. Generaliter ergo si ponatur $(1+az+\mathfrak{b}z^2+\gamma z^3+\delta z^4+\&c.)^{m-1} = 1+Az+Bz^2+Cz^3+Dz^4+Ez^5+\&c.$, hujus Seriei singuli termini ita ex praecedentibus definientur, ut fit $A = \frac{m-1 [TRUNCATED]