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

HAPPY NEW YEAR𝚺_(n=1) ^9 n^3 !

$$\boldsymbol{\mathcal{HAPPY}}\:\:\boldsymbol{\mathcal{NEW}}\:\:\boldsymbol{\mathcal{YEAR}}\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\mathrm{9}} {\boldsymbol{\sum}}{n}}^{\mathrm{3}} \:! \\ $$

Commented by ajfour last updated on 01/Jan/25

Happy new Year ••25. lets all be 25y old. right age equal year.

$${Happy}\:{new}\:{Year}\:\bullet\bullet\mathrm{25}.\:{lets}\:{all}\:{be}\:\mathrm{25}{y} \\ $$$${old}.\:{right}\:{age}\:{equal}\:{year}. \\ $$

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

$$\underset{{n}=\mathrm{1}\rightarrow{b}} {\overset{\mathrm{9}\rightarrow{a}} {\sum}}\:{n}^{\mathrm{3}} \\ $$$$=\left\{\left({a}+{b}\right)^{\mathrm{3}} +\left({a}−{b}\right)^{\mathrm{3}} \right\}+\left({a}−{b}\right)^{\mathrm{3}} +{b} \\ $$$$=\left\{\left(\mathrm{9}+\mathrm{1}\right)^{\mathrm{3}} +\left(\mathrm{9}−\mathrm{1}\right)^{\mathrm{3}} \right\}+\left(\mathrm{9}−\mathrm{1}\right)^{\mathrm{3}} +\mathrm{1} \\ $$$$=\left\{\mathrm{10}^{\mathrm{3}} +\mathrm{8}^{\mathrm{3}} \right\}+\mathrm{8}^{\mathrm{3}} +\mathrm{1} \\ $$$$=\mathrm{1000}+\mathrm{512}+\mathrm{513} \\ $$$$=\mathrm{2025}! \\ $$$${H}\Lambda\rho\rho\gamma\:\:\eta{e}\omega\:\gamma{e}\alpha\pi\:!\: \\ $$$$\:\:\left(\:^{\bullet} \underset{\smile} {\mho}^{\bullet} \:\right) \\ $$$${DON}\:'\:{T}\:{DO}\:{FACTORIAL}\:{BY}\:{MISTAKE} \\ $$

Answered by golsendro last updated on 31/Dec/24

Σ_(n=1) ^9 n^3 = (Σ_(n=1) ^9 n)^2 =45^2 =2025

$$\:\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}\mathrm{n}^{\mathrm{3}} =\:\left(\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}\mathrm{n}\right)^{\mathrm{2}} =\mathrm{45}^{\mathrm{2}} =\mathrm{2025} \\ $$

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

Σ_(n=1) ^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♥)!!!!

$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}{n}^{\mathrm{3}} =\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} +\mathrm{5}^{\mathrm{3}} +\mathrm{6}^{\mathrm{3}} +\mathrm{7}^{\mathrm{3}} +\mathrm{8}^{\mathrm{3}} +\mathrm{9}^{\mathrm{3}} \\ $$$$=\mathrm{1}+\mathrm{8}+\mathrm{27}+\mathrm{64}+\mathrm{125}+\mathrm{126}+\mathrm{343}+\mathrm{512}+\mathrm{729} \\ $$$$=\mathrm{2025}\left(\mathrm{Happy}\:\mathrm{new}\:\mathrm{year}\heartsuit\right)!!!! \\ $$

Answered by MathematicalUser2357 last updated on 31/Dec/24

2025

$$\mathrm{2025} \\ $$

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