משפט: סכום ריבועים (Square pyramidal number) הינו ∑ k = 1 n k 2 = n ( n + 1 ) ( 2 n + 1 ) 6 {\displaystyle \sum _{k=1}^{n}k^{2}={\frac {n(n+1)(2n+1)}{6}}}
הוכחה:
בסיס: ∑ k = 1 1 k 2 = 1 ( 1 + 1 ) ( 2 + 1 ) 6 {\displaystyle \sum _{k=1}^{1}k^{2}={\frac {1(1+1)(2+1)}{6}}}
מכיוון ראשון, ∑ k = 1 1 k 2 = 1 2 = 1 {\displaystyle \sum _{k=1}^{1}k^{2}=1^{2}=1}
מכיוון שני, 1 ( 1 + 1 ) ( 2 + 1 ) 6 = 6 6 = 1 {\displaystyle {\frac {1(1+1)(2+1)}{6}}={\frac {6}{6}}=1}
נניח כי הטענה נכונה לכל n {\displaystyle n} : ∑ k = 1 n k 2 = n ( n + 1 ) ( 2 n + 1 ) 6 {\displaystyle \sum _{k=1}^{n}k^{2}={\frac {n(n+1)(2n+1)}{6}}}
נוכיח את נכונותה לכל n + 1 {\displaystyle n+1} : ∑ k = 1 n + 1 k 2 = ( n + 1 ) ( n + 2 ) ( 2 n + 2 + 1 ) 6 {\displaystyle \sum _{k=1}^{n+1}k^{2}={\frac {(n+1)(n+2)(2n+2+1)}{6}}}
מכיוון אחד, ∑ k = 1 n + 1 k 2 = ∑ k = 1 n k 2 + ( n + 1 ) 2 {\displaystyle \sum _{k=1}^{n+1}k^{2}=\sum _{k=1}^{n}k^{2}+(n+1)^{2}}
על פי הנחת האינדוקציה ∑ k = 1 n k 2 ⏟ n ( n + 1 ) ( 2 n + 1 ) 6 + ( n + 1 ) 2 = ( n + 1 ) ( n + 2 ) ( 2 n + 2 + 1 ) 6 {\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}}}
n ( n + 1 ) ( 2 n + 1 ) 6 + ( n + 1 ) 2 = ( n + 1 ) ( n + 2 ) ( 2 n + 2 + 1 ) 6 {\displaystyle {\frac {n(n+1)(2n+1)}{6}}+(n+1)^{2}={\frac {(n+1)(n+2)(2n+2+1)}{6}}}
n ( n + 1 ) ( 2 n + 1 ) + 6 ( n 2 + 2 n + 1 ) = ( n 2 + 2 n + n + 2 ) ( 2 n + 3 ) {\displaystyle n(n+1)(2n+1)+6(n^{2}+2n+1)=(n^{2}+2n+n+2)(2n+3)}
n ( 2 n 2 + 3 n + 1 ) + 6 n 2 + 12 n + 6 = 2 n 3 + 3 n 2 + 4 n 2 + 6 n + 2 n 2 + 3 n + 4 n + 6 {\displaystyle n(2n^{2}+3n+1)+6n^{2}+12n+6=2n^{3}+3n^{2}+4n^{2}+6n+2n^{2}+3n+4n+6}
2 n 3 + 3 n 2 + n + 6 n 2 + 12 n + 6 = 2 n 3 + 3 n 2 + 4 n 2 + 6 n + 2 n 2 + 3 n + 4 n + 6 {\displaystyle 2n^{3}+3n^{2}+n+6n^{2}+12n+6=2n^{3}+3n^{2}+4n^{2}+6n+2n^{2}+3n+4n+6}
n + 6 n 2 + 12 n = 4 n 2 + 6 n + 2 n 2 + 3 n + 4 n {\displaystyle n+6n^{2}+12n=4n^{2}+6n+2n^{2}+3n+4n}