證明：因為Sn=1+2²+3²+.+n²
當n=1時,S1=1代入Sn=n(n+1)(2n+1)1/6 顯然成立
假設當n=k時,Sk=1+2²+3²+.+k²=k(k+1)(2k+1)/6成立
則當n=k+1時,
S（k+1)=1+2²+3²+.+k²+(k+1)²
=Sk+(k+1)²
=k(k+1)(2k+1)/6+(k+1)²
=(k+1)[k(2k+1)/6+k+1]
=(k+1)(2k²+7k+6)/6=(k+1)(k+2)(2k+3)/6
=(k+1)(k+2)[2(k+1)+1]/6
于是當n=k+1時,S（k+1)=(k+1)(k+2)[2(k+1)+1]/6也成立
所以對一切正整數(shù)n,Sn=n(n+1)(2n+1)1/6成立.