"Counting in their heads" - 1895 oil painting

[Albert Chan]

(08-11-2020 12:33 AM)Gerson W. Barbosa Wrote:  It looks like we can proceed to the next level:

\(
\frac{{10}^{3}+{11}^{3}+{12}^{3}+{13}^{3}+{14}^{3}}{450}
\)

Sum of cubes = squared triangular number = \(\binom{n+1}{2}^2\)

S = (10³+11³+12³+13³+14³) = \(\binom{15}{2}^2 - \binom{10}{2}^2\) = (105-45)*(105+45) = 60*150 = 9000

→ S/450 = 20

---

We can rewrite formula in terms of central element, c = a + (n-1)/2

(x+y)³ + (x-y)³ = (x³+3x²y+3xy²+y³) + (x³-3x²y+3xy²-y³) = 2x³+6xy²

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)

S = 12*5*(12² + (4*6)/4) = 60*150 = 9000