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

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