${\displaystyle {}n=1}$

${\displaystyle {}{\begin{aligned}0&=\sum _{j=0}^{1}{\binom {2}{j}}B_{j}\\&=B_{0}+2B_{1}\\&=1+2B_{1},\end{aligned}}}$

also ist ${\displaystyle {}B_{1}=-{\frac {1}{2}}}$.

${\displaystyle {}n=2}$

${\displaystyle {}{\begin{aligned}0&=\sum _{j=0}^{2}{\binom {3}{j}}B_{j}\\&=B_{0}+3B_{1}+3B_{2}\\&=1-{\frac {3}{2}}+3B_{2}\\&=-{\frac {1}{2}}+3B_{2},\end{aligned}}}$

also ist ${\displaystyle {}B_{2}={\frac {1}{6}}}$.

${\displaystyle {}n=3}$

${\displaystyle {}{\begin{aligned}0&=\sum _{j=0}^{3}{\binom {4}{j}}B_{j}\\&=B_{0}+4B_{1}+6B_{2}+4B_{3}\\&=1-2+1+4B_{3}\\&=4B_{3},\end{aligned}}}$

also ist ${\displaystyle {}B_{3}=0}$.

${\displaystyle {}n=4}$

${\displaystyle {}{\begin{aligned}0&=\sum _{j=0}^{4}{\binom {5}{j}}B_{j}\\&=B_{0}+5B_{1}+10B_{2}+10B_{3}+5B_{4}\\&=1-{\frac {5}{2}}+{\frac {10}{6}}+5B_{4}\\&={\frac {6-15+10}{6}}+5B_{4}\\&={\frac {1}{6}}+5B_{4},\end{aligned}}}$

also ist ${\displaystyle {}B_{4}=-{\frac {1}{30}}}$.

${\displaystyle {}n}$

${\displaystyle {}B_{0},B_{1},\ldots ,B_{n-1}}$

${\displaystyle {}\sum _{j=0}^{n}{\binom {n+1}{j}}B_{j}=0\,}$

schreiben wir als

${\displaystyle {}{\binom {n+1}{n}}B_{n}=-{\left(\sum _{j=0}^{n-1}{\binom {n+1}{j}}B_{j}\right)}\,}$

bzw. als

${\displaystyle {}B_{n}=-{\frac {\sum _{j=0}^{n-1}{\binom {n+1}{j}}B_{j}}{\binom {n+1}{n}}}\,.}$

Da die Binomialkoeffizienten natürliche Zahlen sind, steht hier wieder eine rationale Zahl.