"Counting in their heads" - 1895 oil painting

[Gerson W. Barbosa]

(08-11-2020 10:39 AM)Albert Chan Wrote:  → sum-of-n-cubes = cn*(c² + (n²-1)/4)

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

That’s equivalent to what I’ve come up with, except that I have introduced a factor to generalize for evenly spaced sequences:

sum-of-n-cubes = nc(c² + d²(n² - 1)/4)

where

n = number of elements in the evenly spaced sequence
c = central element
d = interval distance

For example,

S = 33³ + 40³ + 47³ + 54³ + 61³ + 68³

n = 6
c = (33 + 68)/2 = 50.5
d = (68 - 33)/(n - 1) = 7

S = 6×50.5(50.5² + 7²(6² - 1)/4) = 902637

Likewise,

sum-of-n-squares = n(c² + d²(n² - 1)/12)

Considering the similarities between both formulae, perhaps they somehow can be generalized for all positive integer powers.