"Counting in their heads"  1895 oil painting

08102020, 03:06 PM
Post: #10




RE: "Counting in their heads"  1895 oil painting
(08092020 08:11 PM)Gerson W. Barbosa Wrote: This should work also for sequences with even number of elements. For instance, "Central element" can be any number, even irrationals. \(s(x) = (x+a)^2 = a^2 + 2ax + x^2 = a^2 + (2a+1)\binom{x}{1} + 2\binom{x}{2}\) For sum of n consecutive squares, starting from a^2: XCas> S(a,n) := [a^2, 2*a+1, 2] * [n, n*(n1)/2, n*(n1)*(n2)/6] XCas> expand(S(a,n)) → n^3/3+a*n^2+a^2*nn^2/2a*n+n/6 S(a,n) = S(0,a+n0)  S(0,a). What happens if we replace 0 by c ? XCas> expand(S(c,a+nc)  S(c,ac)) → n^3/3+a*n^2+a^2*nn^2/2a*n+n/6 c got *eliminated*, giving same expression as S(a,n) For exact center, c = a+(n1)/2, we got a nice compact formula XCas> simplify(S(c(n1)/2, n)) → (12*c^2*n+n^3n)/12  Since we can pick any c, lets try summing T = 88² + 89² + ... + 115², with c=100 [c^2, 2*c+1, 2] = [10000, 201, 2] \([\binom{16}{1}, \binom{16}{2}, \binom{16}{3}]  [\binom{12}{1}, \binom{12}{2}, \binom{12}{3}]\) = [16,120,560]  [12,78,364] = [28,42,924] T = [10000, 201, 2] • [28,42,924] = 280000 + 8442 + 1848 = 290290 Confirm above with exact center: c = (88+115)/2 = 203/2, n = 11588+1 = 28 T = c²n + (n+1)(n)(n1)/12 = (203/2)²*28 + 29*28*27/12 = 203²*7 + 29*7*9 = (41209 + 261)*7 = 290290 

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