"Counting in their heads" - 1895 oil painting
08-11-2020, 03:06 PM
Post: #17
 Albert Chan Senior Member Posts: 2,493 Joined: Jul 2018
RE: "Counting in their heads" - 1895 oil painting
(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
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