Introductio in analysin infinitorum

PER SERIES INFINITAS. 57 CAP. IV. tasse $(1+Z)^{\frac{m-1}{1}} = 1 + \frac{m-1}{1} Z + \frac{(m-1)(m-2)}{1 \cdot 2} Z^2 + \frac{(m-1)(m-2)(m-3)}{1 \cdot 2 \cdot 3} Z^3 + \&c.$ 73. Sit igitur primum $Z = \alpha z$, eritque $(1+\alpha z)^{\frac{m-1}{1}} = 1 + \frac{m-1}{1} \alpha z + \frac{(m-1)(m-2)}{1 \cdot 2} \alpha^2 z^2 + \frac{(m-1)(m-2)(m-3)}{1 \cdot 2 \cdot 3} \alpha^3 z^3 + \&c.$ Scribatur pro hac Serie ista forma generalis $1+Az+Bz^2+Cz^3+ \dots + Mz^{n-1} + Nz^n + \&c.$ atque quilibet coëfficiens $N$ ex præcedente $M$ ita determinabitur ut fit $N = \frac{m-n}{n} \alpha M$. Sic, posito $n=1$, cum fit $M=1$, erit $N=A=\frac{m-1}{1} \alpha$; tum facto $n=2$, ob $M=A=\frac{m-1}{1} \alpha$, erit $N=B=\frac{m-2}{2} \alpha M = \frac{(m-1)(m-2)}{1 \cdot 2} \alpha^2$; similique modo porro $C = \frac{m-3}{3} \alpha B = \frac{(m-1)(m-2)(m-3)}{1 \cdot 2 \cdot 3} \alpha^3$, uti Series ante inventa declarat. 74. Sit $Z = \alpha z + \zeta z^2$, eritque $(1+\alpha z + \zeta z^2)^{\frac{m-1}{1}} = 1 + \frac{m-1}{1} (\alpha z + \zeta z^2) + \frac{(m-1)(m-2)}{1 \cdot 2} (\alpha z + \zeta z^2)^2 + \&c.$ Quod si ergo termini secundum potestates ipsius $z$ disponantur erit $(1+\alpha z + \zeta z^2)^{\frac{m-1}{1}} = 1 + \frac{m-1}{1} \alpha z + \frac{(m-1)(m-2)}{1 \cdot 2} \alpha^2 z^2 + \frac{(m-1)(m-2)(m-3)}{1 \cdot 2 \cdot 3} \alpha^3 z^3 + \&c. + \frac{m-1}{1} \zeta z^2 + \frac{(m-1)(m-2)}{1 \cdot 2} 2 \alpha \zeta z^3 + \&c.$ Euleri Introduct. in Anal. infin. parv. H Scri-