Introductio in analysin infinitorum

RECURRENTIBUS. 189 CAP. XIII. $B=+\frac{1}{4}$, unde fit terminus generalis $= (-\frac{1}{4} \sin \frac{1}{2} (n+1)\omega + \frac{1}{4} \sin \frac{1}{2} n\omega)z^n$. Quare colligendo erit terminus generalis quaesitus $= (\frac{3}{4}n + \frac{9}{8})z^n \pm \frac{1}{8}z^n - \frac{1}{4}(\sin \frac{1}{2}(n+1)\omega - \sin \frac{1}{2}n\omega)z^n$. Hinc si fuerit $n=4m+0$ $n=4m+1$ $n=4m+2$ $n=4m+3$ erit terminus generalis $(\frac{3}{4}n+1)z^n$ $(\frac{3}{4}n+\frac{5}{4})z^n$ $(\frac{3}{4}n+\frac{3}{2})z^n$ $(\frac{3}{4}n+\frac{3}{4})z^n$ Ita, si $n=50$, valebit $n=4m+2$, eritque terminus $= 39250$. 224 [TRUNCATED]