I could not really even find any apparent formula.
2007-04-07 14:37:33 · 9 answers · asked by Edgard L 2 in Science & Mathematics ➔ Mathematics
not a series of cubes. there is no straight formula. there are 3rd and 4th power exponents involved, and they are not even in order
2007-04-07 14:54:19 · answer #1 · answered by Anonymous · 1⤊ 0⤋
This is NOT a series of cubes!
256 = 2^8 and 625 = 5^4.
The sum is 1^3 + 2^3 + 3^3 + 6^3 + 2^8 + 5^4.
I don't think one formula will cover it.
2007-04-07 14:49:07 · answer #2 · answered by steiner1745 7 · 1⤊ 1⤋
the respond is sixty 4. a million cubed is a million, 2 cubed is 8, 3 cubed is 27, 4 cubed is sixty 4, 5 cubed is a hundred twenty five, 6 cubed is 216. commencing with the 1st type interior the sequence being the 1st term, it may desire to be defined as n^3 = f(n)
2016-12-08 21:09:35 · answer #3 · answered by ? 4 · 0⤊ 0⤋
The answer is 1133.
1^3 + 2^3 + 3^3 + 4^4 + 5^4.
If the series is not infinite, then you could come up with any criteria for why the power changes from 3 to 4.
2007-04-07 14:45:33 · answer #4 · answered by Dr D 7 · 0⤊ 2⤋
1133
2007-04-07 22:56:23 · answer #5 · answered by Anonymous · 0⤊ 1⤋
It's a series of cubes:
(1^3) + (2^3) + (3^3) + .......
Thus the formula is [sum from n=1 to infinity] (n^3)
2007-04-07 14:43:27 · answer #6 · answered by poorcocoboiboi 6 · 0⤊ 3⤋
The sum of x for x =1 to 6 of
f(x)=x^3.
don't know how to make the sigma sign with the upper and lower limit around it, but I'm sure oyu know what i mean..
Since your list is finite, its not infinite, its only six members of the set.
2007-04-07 14:43:58 · answer #7 · answered by The Big Lebowski 3 · 0⤊ 2⤋
It's 1123. (Not even good at math either and i didn't use a calculator either.)
2007-04-07 14:42:48 · answer #8 · answered by audreyjeromin 2 · 0⤊ 2⤋
I know if you add them together you get 1,133!
2007-04-07 14:42:20 · answer #9 · answered by Kiwi 4 · 0⤊ 3⤋