HAPPY NEW YEAR𝚺_(n=1) ^9 n^3 !


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Question Number 215200 by mr W last updated on 31/Dec/24

$HAPPY NEW YEAR {\underset{n = 1}{\sum^{9} n}}^{3}!$
Commented by ajfour last updated on 01/Jan/25

$H a p p y n e w Y e a r ∙ ∙ 25 . l e t s a l l b e 25 y$ $o l d . r i g h t a g e e q u a l y e a r .$
	Answered by Marzuk last updated on 31/Dec/24
	$Σ_(n=1→b) ^(9→a) n^3 ={(a+b)^3 +(a−b)^3 }+(a−b)^3 +b ={(9+1)^3 +(9−1)^3 }+(9−1)^3 +1 ={10^3 +8^3 }+8^3 +1 =1000+512+513 =2025! HΛρργ ηeω γeαπ ! (^• ℧^• _⌣ ) DON ′ T DO FACTORIAL BY MISTAKE$
	$\underset{n = 1 \to b}{\sum^{9 \to a}} n^{3}$ $= {{(a + b)}^{3} + {(a - b)}^{3}} + {(a - b)}^{3} + b$ $= {{(9 + 1)}^{3} + {(9 - 1)}^{3}} + {(9 - 1)}^{3} + 1$ $= {10^{3} + 8^{3}} + 8^{3} + 1$ $= 1000 + 512 + 513$ $= 2025!$ $H Λ ρ ρ γ η e ω γ e α π!$ $(^{∙} {\underset{⌣}{℧}}^{∙})$ $D O N^{'} T D O F A C T O R I A L B Y M I S T A K E$
	Answered by golsendro last updated on 31/Dec/24

	$\underset{n = 1}{\sum^{9}} n^{3} = {(\underset{n = 1}{\sum^{9}} n)}^{2} = 45^{2} = 2025$
	Commented by ajfour last updated on 01/Jan/25
	45 is my present age. ha ha. Feel o m at the root of time itself.
	Answered by MrGaster last updated on 31/Dec/24

	$\underset{n = 1}{\sum^{9}} n^{3} = 1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} + 6^{3} + 7^{3} + 8^{3} + 9^{3}$ $= 1 + 8 + 27 + 64 + 125 + 126 + 343 + 512 + 729$ $= 2025 (Happy new year ♡)!!!!$
	Answered by MathematicalUser2357 last updated on 31/Dec/24

	$2025$
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