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1^3 - 2^3 + 3^3 - 4^3 ... + 99^3 - 100^3 = ?
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1^3 - 2^3 + 3^3 - 4^3 ... + 99^3 - 100^3 = ?
FLOUR
04/24/2025 04:43:45 PM
04/24/2025
04:43 PM
#1
Pro tip reminder: 1^2 + 2^2 + 3^2 + 4^2 ... +n^2 = n(n+1)(2n+1)/6
The alternate domination of one faction over another is itself a frightful despotism. - George Washington
DrizztLink
04/24/2025 04:44:02 PM
04/24/2025
04:44 PM
#2
Pretty sure it doesn't equal ?.
He/Him http://guidesmedia.ign.com/guides/9846/images/slowpoke.gif https://i.imgur.com/M8h2ATe.png
https://i.imgur.com/6ezFwG1.png
Princess_Eev
04/24/2025 04:44:54 PM
04/24/2025
04:44 PM
#3
https://math.stackexchange.com/questions/1576214/how-to-find-alternating-sum-of-powers
(she/her) | Math nerd, lover of pineapple on pizza <3
Infodumps:
Dark Cloud 2: https://is.gd/PSE06J | Pokemon MD: https://is.gd/1lK5bM
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1^3 - 2^3 + 3^3 - 4^3 ... + 99^3 - 100^3 = ?
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