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Wir betrachten die naürliche Additionstabelle bis zu einer bestimmten Zahl ${\displaystyle {}n}$, also

${\displaystyle {}+}$ ${\displaystyle {}1}$ ${\displaystyle {}2}$ ${\displaystyle {}3}$ ${\displaystyle {}\ldots }$ ${\displaystyle {}n-1}$ ${\displaystyle {}n}$
${\displaystyle {}1}$ ${\displaystyle {}1+1}$ ${\displaystyle {}1+2}$ ${\displaystyle {}1+3}$ ${\displaystyle {}\ldots }$ ${\displaystyle {}1+n-1}$ ${\displaystyle {}1+n}$
${\displaystyle {}2}$ ${\displaystyle {}2+1}$ ${\displaystyle {}2+2}$ ${\displaystyle {}2+3}$ ${\displaystyle {}\ldots }$ ${\displaystyle {}2+n-1}$ ${\displaystyle {}2+n}$
${\displaystyle {}3}$ ${\displaystyle {}3+1}$ ${\displaystyle {}3+2}$ ${\displaystyle {}3+3}$ ${\displaystyle {}\ldots }$ ${\displaystyle {}3+n-1}$ ${\displaystyle {}3+n}$
${\displaystyle {}\ldots }$ ${\displaystyle {}\vdots }$ ${\displaystyle {}\vdots }$ ${\displaystyle {}\vdots }$ ${\displaystyle {}\vdots }$ ${\displaystyle {}\vdots }$ ${\displaystyle {}\vdots }$
${\displaystyle {}n-1}$ ${\displaystyle {}n-1+1}$ ${\displaystyle {}n-1+2}$ ${\displaystyle {}n-1+3}$ ${\displaystyle {}\ldots }$ ${\displaystyle {}n-1+n-1}$ ${\displaystyle {}n-1+n}$
${\displaystyle {}n}$ ${\displaystyle {}n+1}$ ${\displaystyle {}n+2}$ ${\displaystyle {}n+3}$ ${\displaystyle {}\ldots }$ ${\displaystyle {}n+n-1}$ ${\displaystyle {}n+n}$

Zeige durch Induktion, dass die Gesamtsumme aller in der Tabelle auftretenden Summen gleich ${\displaystyle {}(n+1)n^{2}}$ ist, also

${\displaystyle {}\sum _{1\leq i\leq n,\,1\leq j\leq n}(i+j)=(n+1)n^{2}\,.}$