From the formula (n+1) 3 = n 3+3n 2+3n+1,there is (n+1) 3-n 3 = 3n 2+3n+1,

n^3-(n- 1)^3=3(n- 1)^2+3(n- 1)+ 1

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2^3- 1^3=3* 1^2+3* 1+ 1

Add up the above n expressions, and then sum them in groups. It is not difficult to find the square of 1 +2+…+N =n(n+ 1)(2n+ 1)/6.

Similarly, using the formula (n+1) 4 = n 4+4n3+6n2+4n+ 1, the square of cube +2 of1can be multiplied by the square of (n+1).