(08-11-2020 12:43 PM)Gerson W. Barbosa Wrote: [ -> ]sum-of-n-squares = n(c² + d²(n² - 1)/12)

sum-of-n-cubes = nc(c² + d²(n² - 1)/4)

Considering the similarities between both formulae, perhaps they somehow can be generalized

for all positive integer powers.

Perhaps this explain the similarity, for S

_{p} = a

^{p} + (a+1)

^{p} + ... + b

^{p}, b = a+n-1

With center c = (a+b)/2, g = (n-1)/2 = b-c = c-a, we have

S(a,n) = S(c,b-c+1) - S(c,a-c)

=

S(c,g+1) - S(c,-g)
= Finite-Difference-Diagonals • \(\left(

[\binom{g+1}{1},\binom{g+1}{2},\cdots] - [\binom{-g}{1},\binom{-g}{2},\cdots] \right)\)

Let's see what the combinatorial coefficients look like:

XCas> g := (n-1)/2

XCas> coef := makelist(r -> simplify(comb(g+1,r) - comb(-g,r)),1,6)

→ \(\Large [n,\,0,\,\frac{(n^{3}-n)}{24},\,\frac{(-n^{3}+n)}{24},\,\frac{(n^{5}+70\cdot n^{3}-71\cdot n)}{1920},\,\frac{(-n^{5}-30\cdot n^{3}+31\cdot n)}{960}]\)

coef(1) = n, which make first term of S

_{p} = c^p*n

coef(2) = 0, which make the central element based formula compact (Δf * 0 = 0)

coef(3) = -coef(4), which meant dot-products have this common factor.

Finite Difference table for (c+x)² and (c+x)³ // note: [1,0,0] ≡ c², [1,2,1] ≡ c²+2c+1

Code:

`x (c+x)^2`

0 [1,0,0]

1 [1,2,1] [2,1]

2 [1,4,4] [2,3] [2]

x (c+x)^3

0 [1,0,0,0]

1 [1,3,3,1] [3,3,1]

2 [1,6,12,8] [3,9,7] [6,6]

3 [1,9,27,27] [3,15,19] [6,12] [6]

If we let k = coef(3) = (n³-n)/24:

sum-of-n-squares = c^2*n + 2k

sum-of-n-cubes = c^3*n + (6c+6)*k + 6*(-k) = c*(c^2*n + 6k)

coef(4) and coef(5) are not close, which meant similarity ends after cubes

XCas> sumpow(p) := expand(simplify(sum(x^p, x=c-g .. c+g)))

XCas> for(p=1; p<6; p++) {print(p, sumpow(p))}

1, c*n

2, (n^3)/12+c^2*n-n/12

3, (c*n^3)/4+c^3*n-(c*n)/4

4, (n^5)/80+(c^2*n^3)/2+c^4*n-(n^3)/24-(c^2*n)/2+(7*n)/240

5, (c*n^5)/16+(5*c^3*n^3)/6+c^5*n-(5*c*n^3)/24-(5*c^3*n)/6+(7*c*n)/48