I believe: sum(i=1,n)i^3 = (n*(n+1)/2)^2 Trivially true for n=1. And if true up to n, then: sum(i=1,n+1)i^3 = sum(i=1,n)i^3 + (n+1)^3 = (n*(n+1)/2)^2 + (n+1)^3 = (1/4)*(n*(n+1))^2 + (n+1)^3 = (1/4) * ((n*(n+1))^2 + 4*(n+1)^3) = (1/4) * (n^2*(n+1)^2 + 4*(n+1)*(n+1)^2) = (1/4) * (n^2 + 4*(n+1))*(n+1)^2 = (1/4) * (n+2)^2*(n+1)^2 = ((n+1)*(n+2)/2)^2 So it's true for n+1. Tom christopher landauer writes:
hihi -
i was under the impression that an initial sum of cubes is always a square - isn't there a summation formula to that effect?
more later,
chris
On 2020-05-09 14:09, rcs@xmission.com wrote:
If we allow the sum to be a different power ... The sum of the first N cubes is the square of a triangle.
1^3 + 2^3 + 3^3 + 4^3 = 10^2
Sometimes a triangle is also a square: T(8) = 36 = 6^2.
The square of a square is a 4th power:
1^3 + 2^3 + ... + 8^3 = 1296 = 36^2 = 6^4
The next triangle-square is 1225.
1^3 + 2^3 + ... + 49^3 = 1225^2 = 35^4
There's a hidden fact in the Ramanujan-Hardy 1729 story, but it's never mentioned. 1729 is a (base 2) pseudoprime: 2^1728 (mod 1729) = 1.
Here's how I flunked my Ramanujan test. My University of Arizona employee number was 41616. I didn't look at it very often-- mainly when I used my ID card to check out a library book. Years later, I noticed the number was special. In fact, very special. Puzzle: Why is it special? Puzzle2: Why is it very special?
Rich
------- Quoting James Buddenhagen <jbuddenh@gmail.com>:
Well, there is 1^1 + 2^1 = 3^1, which I realize does not have k > 2. Closely related to your question, so perhaps of interest even though it is not a match is: http://oeis.org/A030052/a030052.txt
Jim
On Fri, May 8, 2020 at 2:18 PM Dan Asimov <dasimov@earthlink.net> wrote:
Famously,
1^2 + 2^2 + 3^2 + ... + 24^2 = 70^2
is the only case where the sum of an initial sequence of squares equals another square (ignoring 1^2 = 1^2).
Are there any non-trivial examples of this for powers higher than 2 ?
(For non-initial cases, 3^3 + 4^3 + 5^3 = 6^3 is the first example, and this website <https://www.mathpages.com/home/kmath147.htm> gives many other examples for cubes.)
But I'm interested in when 1^k + 2^k + ... + n^k is an exact kth power for k > 2 and of course n > 1.
?Dan