Prove:

1。 When n= 1, 2+3+4= 1+8, and the equation holds.

2。 Let n = k> When =2, the equation holds, then (k 2+1)+(k 2+2)+...+(k+1) 2 = k 3+(k+1) 3.

That is, (K2+1)+(K2+2)+...+(K2+2k+1) = K3+(K+1) 3.

For n=k+ 1, there is

[(k+ 1)^2+ 1]+[(k+ 1)^2+2]+...+(k+2)^2

=[(k+ 1)^2+ 1]+[(k+ 1)^2+2]+...+[(k+ 1)^2+2k+3]

=[(k^2+ 1)+(2k+ 1)]+[(k^2+2)+(2k+ 1)]+...+[(k^2+2k+ 1)+(2k+ 1)]+[(k^2+2k+2)+(2k+ 1)]+[(k^2+2k+3)+(2k+ 1)]

=k^2+ 1)+(k^2+2)+...+(k^2+2k+ 1)+(2k+ 1)(2k+ 1)+[(k^2+2k+2)+(2k+ 1)]+[(k^2+2k+3)+(2k+ 1)]

=k^3+(k+ 1)^3+(2k+ 1)(2k+ 1)+[(k^2+2k+2)+(2k+ 1)]+[(k^2+2k+3)+(2k+ 1)]

=k^3+(k+ 1)^3+6k^2+ 12k+8

=(k+ 1)^3+(k+2)^3

The equation pair n=k+ 1 also holds.

3。 Finally, this equation applies to all positive integers n.