"Counting in their heads"  1895 oil painting

08112020, 03:06 PM
Post: #17




RE: "Counting in their heads"  1895 oil painting
(08112020 10:39 AM)Albert Chan Wrote: Sum of cubes = squared triangular number = \(\binom{n+1}{2}^2\) Proof: f(x) = x³ = (x³x) + x = (x+1)^{3} + x^{1} F(x) = (x+1)^{4}/4 + x^{2}/2 = \(\binom{x}{2} × {(x+1)(x2) + 2 \over 2} = \binom{x}{2}^2\) sumofncubes = F(n+1) = \(\binom{n+1}{2}^2\) (08112020 10:39 AM)Albert Chan Wrote: sumofncubes \(= c^3 n + 6c×(1^2 + 2^2 + \cdots + ({n1\over2})^2) Although above formula assumed (n1)/2 is integer, this is not necessary. XCas> S(a,n) := simplify(sum((a+x)^3, x=0..n1)) XCas> factor(S(c(n1)/2,n)) → n*c*(4*c^2+n^21)/4 Again, any "central element" work, with differences back to S(a,n) XCas> simplify(S(c,a+nc)  S(c,ac)  S(a,n)) → 0 Redo previous example, using simple sumofncubes 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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