1+2*2+3*3+...+n*n=1+1*2+2+2*3+3+3*4+4+.(n-1)n+n=(1+2+3+...+n)+[1*2+2*3+3*4+.(n-1)*n]=n(n+1)/2+[1*2*3+2*3*3+3*4*3+.(n-1)*n*3]/3=n(n+1)/2+[1*2*3+2*3*4-1*2*3+3*4*5-2*3*4+.(n-1)n(n+1)-(n-2)(n-1)n]/3=n(n+1)/2+(n-1)n(n+1)/3=n(n+1)(2n+1)/6