Cubic difference formula: a 3-b 3 = (a-b) (a 2+ab+b 2)

Formula extension:13+2 3+... n 3 = [n (n+1)/2] 2 = (1+2+...+n) 2.

Formula proof:

We know:

The summation formula of the sum of the zeroth power ∑ n 0 = n is10+20+...+n 0 = n.

The sum formula of 1 power sum is ∑ n 1 = n (n+ 1)/2, that is,11+21+...+n1= n.

The sum formula of square sum ∑ n 2 = n (n+1)/6 is12+...+n 2 = n (n+1) (2n+65438).

Take the formula: (x+1) 4-x4 = 4 * x3+6 * x2+4 * x+1.

The coefficient can be determined by Yang Hui triangle.

Then there is:

(n+ 1)^4-n^4=4n^3+6n^2+4n+ 1....................................( 1)

n^4-(n- 1)^4=4(n- 1)^3+6(n- 1)^2+4(n- 1)+ 1.......................(2)

(n- 1)^4-(n-2)^4=4(n-2)^3+6(n-2)^2+4(n-2)+ 1..................(3)

...................

2 4-14 = 4×13+6×12+4×1+/................................... (noun).

So (1)+(2)+(3)+...+(n) Yes.

Left = (n+ 1) 4- 1

Right = 4 ( 1 3+2 3+3 3+...+n 3)+6 ( 1 2+2 2+3 2+...+n ^ 2)+4(65438+)。

therefore

Substitute the above three proven formulas.

4( 1^3+2^3+3^3+......+N^3)+6( 1^2+2^2+3^2+......+N^2)+4( 1+2+3+......+N)+N=(N+ 1)^4- 1

The result is 4 (13+23+33+n 3)+n (n+1) (2n+1)+2n (n+1)+n = n.

The shifted term is13+23+3+...+n3 =1/4 (n4+4n3+6n2+4n-n-2n2-2n-2n3-3n

1 3+2 3+3 3+ ... After the similar items on the right of the equal sign are merged,+n 3 =1/4 (n 4+2n 3+n 2).

that is

1^3+2^3+3^3+......+N^3= 1/4 [N(N+ 1)]^2

The cubic sum formula is derived.

1^3+2^3+3^3+......+N^3= 1/4 [N(N+ 1)]^2

I looked it up online, not by myself. . .