1^2+2^2+...+n^2=n(n+1)(2n+1)/6
n=1,略
假設(shè)n=k成立
1^2+2^2+...+k^2=k(k+1)(2k+1)/6
則n=k+1
1^2+2^2+...+k^2+(k+1)^2
=k(k+1)(2k+1)/6+(k+1)^2
=(k+1)[k(2k+1)+6(k+1)]/6
=(k+1)[2k^2+7k+6)/6
=(k+1)(k+2)(2k+3)/6
=(k+1)[(k+1)+1][2(k+1)+1]/6
綜上
1^2+2^2+...+n^2=n(n+1)(2n+1)/6