משפט: סכום ריבועים (Square pyramidal number) הינו ${\displaystyle \sum _{k=1}^{n}k^{2}={\frac {n(n+1)(2n+1)}{6}}}$

הוכחה:

בסיס: ${\displaystyle \sum _{k=1}^{1}k^{2}={\frac {1(1+1)(2+1)}{6}}}$

מכיוון ראשון, ${\displaystyle \sum _{k=1}^{1}k^{2}=1^{2}=1}$

מכיוון שני, ${\displaystyle {\frac {1(1+1)(2+1)}{6}}={\frac {6}{6}}=1}$

נניח כי הטענה נכונה לכל ${\displaystyle n}$ :${\displaystyle \sum _{k=1}^{n}k^{2}={\frac {n(n+1)(2n+1)}{6}}}$

נוכיח את נכונותה לכל ${\displaystyle n+1}$: ${\displaystyle \sum _{k=1}^{n+1}k^{2}={\frac {(n+1)(n+2)(2n+2+1)}{6}}}$

מכיוון אחד, ${\displaystyle \sum _{k=1}^{n+1}k^{2}=\sum _{k=1}^{n}k^{2}+(n+1)^{2}}$

על פי הנחת האינדוקציה ${\displaystyle \underbrace {\sum _{k=1}^{n}k^{2}} _{\frac {n(n+1)(2n+1)}{6}}+(n+1)^{2}={\frac {(n+1)(n+2)(2n+2+1)}{6}}}$

${\displaystyle {\frac {n(n+1)(2n+1)}{6}}+(n+1)^{2}={\frac {(n+1)(n+2)(2n+2+1)}{6}}}$

${\displaystyle n(n+1)(2n+1)+6(n^{2}+2n+1)=(n^{2}+2n+n+2)(2n+3)}$

${\displaystyle n(2n^{2}+3n+1)+6n^{2}+12n+6=2n^{3}+3n^{2}+4n^{2}+6n+2n^{2}+3n+4n+6}$

${\displaystyle 2n^{3}+3n^{2}+n+6n^{2}+12n+6=2n^{3}+3n^{2}+4n^{2}+6n+2n^{2}+3n+4n+6}$

${\displaystyle n+6n^{2}+12n=4n^{2}+6n+2n^{2}+3n+4n}$