Hey guys!

Leap

New Member
Oct 6, 2020
2
2
3
Canada
Hey guys!

It's Leap, just hopped on to the forums to say how much of a wonderful thing this community is. ๐Ÿ‘‹

Also, I posted earlier in the discord general chat that if you add numbers up to n, and squared the sum, you can get the same number by adding the individual cubes of each of those numbers.
ie (1 + 2 + 3 + 4 + 5)^2 is 225, and so is 1^3 + 2^3 + 3^3 + 4^3 + 5^3

Well, there's even more!
Any square number can be written as the sum of consecutive odd numbers! (The last term being the sum of consecutive numbers * 2 - 1) So 225 (the square of 1+2+3+4+5) can be written also as (1 + 3 + 5 + ... + 29)! That means every set of (1 + 2 + 3 + .. + n)^2 equals that AND two other sets of consecutive sums! (1^3 + 2^3 + ... + n^3 and 1 + 3 + ... + m where m = (1+2+3+...+n)*2-1)

Thank you for stopping by to read this! Hope you all have a good day today!
 
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VashiepUwU

New Member
Admin+
Mar 1, 2020
46
6
8
Michigan, USA
Hey guys!

It's Leap, just hopped on to the forums to say how much of a wonderful thing this community is. ๐Ÿ‘‹

Also, I posted earlier in the discord general chat that if you add numbers up to n, and squared the sum, you can get the same number by adding the individual cubes of each of those numbers.
ie (1 + 2 + 3 + 4 + 5)^2 is 225, and so is 1^3 + 2^3 + 3^3 + 4^3 + 5^3

Well, there's even more!
Any square number can be written as the sum of consecutive odd numbers! (The last term being the sum of consecutive numbers * 2 - 1) So 225 (the square of 1+2+3+4+5) can be written also as (1 + 3 + 5 + ... + 29)! That means every set of (1 + 2 + 3 + .. + n)^2 equals that AND two other sets of consecutive sums! (1^3 + 2^3 + ... + n^3 and 1 + 3 + ... + m where m = (1+2+3+...+n)*2-1)

Thank you for stopping by to read this! Hope you all have a good day today!
You definitely understand the language of the universe better than I do.

Welcome :cool:
 

Tensei

New Member
Admin+
Jan 7, 2020
48
19
58
hello leap! my favorite math teacher. hope u are doing well. you should apply for member.
 
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