"Counting in their heads" - 1895 oil painting

[Albert Chan]

(08-11-2020 10:39 AM)Albert Chan Wrote:  Sum of cubes = squared triangular number = \(\binom{n+1}{2}^2\)

Proof:

f(x) = x³ = (x³-x) + x = (x+1)³ + x¹
F(x) = (x+1)⁴/4 + x²/2 = \(\binom{x}{2} × {(x+1)(x-2) + 2 \over 2} = \binom{x}{2}^2\)

sum-of-n-cubes = F(n+1) = \(\binom{n+1}{2}^2\)

(08-11-2020 10:39 AM)Albert Chan Wrote:  sum-of-n-cubes \(= c^3 n + 6c×(1^2 + 2^2 + \cdots + ({n-1\over2})^2)
= c^3 n + 6c × \binom{n+1}{3}/4 \)

→ sum-of-n-cubes = cn*(c² + (n²-1)/4)

Although above formula assumed (n-1)/2 is integer, this is not necessary.

XCas> S(a,n) := simplify(sum((a+x)^3, x=0..n-1))
XCas> factor(S(c-(n-1)/2,n))             → n*c*(4*c^2+n^2-1)/4

Again, any "central element" work, with differences back to S(a,n)

XCas> simplify(S(c,a+n-c) - S(c,a-c) - S(a,n))         → 0

Redo previous example, using simple sum-of-n-cubes formula (c=1)

S = 33³ + 40³ + 47³ + 54³ + 61³ + 68³
    = 7³ * ((33/7)³ + (33/7+1)³ + ... + (68/7)³)
    = \(7^3 \left(\binom{75/7}{2}^2 - \binom{33/7}{2}^2) \right)\)
    = \(\Large{(75×68)^2 - (33×26)^2 \over 7×4}\)
    = 902637