"Counting in their heads" - 1895 oil painting
08-10-2020, 12:49 PM (This post was last modified: 08-10-2020 02:38 PM by Albert Chan.)
Post: #9
 Albert Chan Senior Member Posts: 2,354 Joined: Jul 2018
RE: "Counting in their heads" - 1895 oil painting
(08-10-2020 11:50 AM)Gerson W. Barbosa Wrote:  $$\frac{{55}^{2}+{65}^{2}+{75}^{2}+{85}^{2}+{95}^{2}+{105}^{2}+{115}^{2}+{125}^{2}​+{135}^{2}+{145}^{2}}{10825}$$

Also easily done, when you know the right formula :-)

$$s(x) = (a+bx)^2 = a^2 + 2abx + b^2x^2 = a² + b(2a+b)\binom{x}{1} + 2b^2\binom{x}{2}$$

→ Σ(s(x), x=0..n-1) = [a², b(2a+b), 2b²] • [nC1, nC2, nC3]

With a=55, b=10, n=10

S = [a², b(2a+b), 2b²] • [nC1, nC2, nC3]
﻿ ﻿ ﻿ = [3025, 1200, 200] • [10, 45, 120]
﻿ ﻿ ﻿ = 30250 + 54000 + 24000
﻿ ﻿ ﻿ = 108250

→ S / 10825 = 10

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We can also use central element, c = a + b(n-1)/2

S = c²n + b²(n+1)(n)(n-1)/12

With c = (55+145)/2 = 100, b=10, n=10:

S = 100²*10 + 10²*11*10*9/12 = 100000 + 8250 = 108250

Or, simply force b=1

(55² + 65² + ... + 145²) / 10825
= (5.5² + 6.5² + ... + 14.5²) / 108.25 ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿// c=n=10
= (1000 + 11*10*9/12) / 108.25
= 10
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