Squared triangular number
The sum of the first n cubes, which equals the square of the nth triangular number
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Squared triangular number
The sum of the first n cubes, which equals the square of the nth triangular number
In number theory, the sum of the first n cubes is the square of the nth triangular number. That is, 1 3 + 2 3 + 3 3 + ⋯ + n 3 = ( 1 + 2 + 3 + ⋯ + n ) 2 . {\displaystyle 1^{3}+2^{3}+3^{3}+\cdots +n^{3}=\left(1+2+3+\cdots +n\right)^{2}.} The same equation may be written more compactly using the mathematical notation for summation: ∑ k = 1 n k 3 = ( ∑ k = 1 n k ) 2 . {\displaystyle \sum _{k=1}^{n}k^{3}=\left(\sum _{k=1}^{n}k\right)^{2}.} This identity is sometimes called Nicomachus's theorem, after Nicomachus of Gerasa (c. 60 – c. 120 CE).
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From Wikipedia
In number theory, the sum of the first n cubes is the square of the nth triangular number. That is, 1 3 + 2 3 + 3 3 + ⋯ + n 3 = ( 1 + 2 + 3 + ⋯ + n ) 2 . {\displaystyle 1^{3}+2^{3}+3^{3}+\cdots +n^{3}=\left(1+2+3+\cdots +n\right)^{2}.} The same equation may be written more compactly using the mathematical notation for summation: ∑ k = 1 n k 3 = ( ∑ k = 1 n k ) 2 . {\displaystyle \sum _{k=1}^{n}k^{3}=\left(\sum _{k=1}^{n}k\right)^{2}.} This identity is sometimes called Nicomachus's theorem, after Nicomachus of Gerasa (c. 60 – c. 120 CE).
Text: Wikipédia, CC BY-SA 4.0. · Image: Cmglee (CC BY-SA 3.0) ·
