求證1²+2²+3²+.+n²==n(n+1)(2n+1)/6
證明：∵(k+1)³=k³+3k²+3k+1
==>k²=[(k+1)³-k³]/3-k-1/3
∴1²=(2³-1³)/3-1-1/3
2²=(3³-2³)/3-2-1/3
3²=(4³-3³)/3-3-1/3
.
n²=[(n+1)³-n³]/3-n-1/3
故1²+2²+3²+.+n²
={[(2³-1³)/3-1-1/3]+[(3³-2³)/3-2-1/3]+[(4³-3³)/3-3-1/3]+.+[[(n+1)³-n³]/3-n-1/3]}
=[(n+1)³-1³]/3-(1+2+3+.+n)-n/3
=n(n²+3n+3)/3-n(n+1)/2-n/3
=n[2(n²+3n+3)-3(n+1)-2]/6
=n(2n²+3n+1)/6
=n(n+1)(2n+1)/6.證畢.