用数学归纳法
①1+2+3+…+n = n(n+1)/2
②验证n=1时，1³ = [1(1+1)/2]² = 1
③令n=k时，1³+2³+3³+…+k³ = k²(k+1)²/4
试证n = k+1时，
1³+2³+3³+…+k³+(k+1)³
= k²(k+1)²/4+(k+1)³
= (k+1)²[k²/4+(k+1)]
= [(k+1)²/4](k+2)²
= (k+1)²[(k+1)+1]²/4
成立。得证。