Hey guys, I just want to share this simple things. I have found pattern of cubic number, and now I want to share it to you.

\(0^{3} = 0\)

\(1^{3} = 1\)

\(2^{3} = 8 = 3 + 5\)

\(3^{3} = 27 = 7 + 9 + 11\)

\(4^{3} = 64 = 13 + 15 + 17 + 19\)

\(5^{3} = 125 = 21 + 23 + 25 +27 + 29\)

\(6^{3} = 216 = 31 + 33 + 35 + 37 + 39 + 41\)

\(7^{3} = 343 = 43 + 45 + 47 + 49 +51 + 53 + 55\)

\(8^{3} = 512 = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71\)

\(9^{3} = 729 = 73 + 75 + 77 + 79 + 81 + 83 + 85 + 87+ 89\)

and so on and on and on !

Do you get the pattern ? Okay, any comments will be appreciated. Thank you...

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## Comments

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TopNewest\(\displaystyle \sum _{ k=0 }^{ n-1 }{ \left( { n }^{ 2 }-n+1+2k \right) ={ n }^{ 3 } }\)

The cube creeps into this thing because of that first term \({ n }^{ 2 }\)

As an example, this is true also, but the sum are quartics.

\(\displaystyle \sum _{ k=0 }^{ n-1 }{ \left( { n }^{ 3 }-n+1+2k \right) ={ n }^{ 4 } }\)

In fact, for any \(a\), this is true

\(\displaystyle \sum _{ k=0 }^{ n-1 }{ \left( { n }^{ a }-n+1+2k \right) ={ n }^{ a+1 } }\)

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Correct me if I am wrong. So, for \(n^{2}\) it will be

\(\sum_{k=0}^{n-1} (n+1+2k) = n^{2}\)

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Actually, leave out \(n\), since \({n}^{1}-n=0\), and we have the classic sum of odd numbers \(1+2k\) to make squares.

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Thanks for formula for ANY power.

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Let me give you a pattern:

\[1^{3} = 1^{2} = 1\]

\[1^{3}+2^{3} = (1+2)^{2} = 9\]

\[1^{3}+2^{3}+3^{3} = (1+2+3)^{2} = 36\]

\[1^{3}+2^{3}+3^{3}+4^{3} = (1+2+3+4)^{2} = 100\]

In general,

\[\boxed{1^{3}+2^{3}+\cdots+n^{3} = (1+2+\cdots+n)^{2}}\]

Can you figure out why this is? ;)

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Sum of cubes = n(n+1)(n)(n+1)/4= [n(n+1)/2]^2= [Sum of the numbers]^2

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In boxed part, \(1^3+2^3\)

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I have edited it! \(\ddot\smile\)

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@Jake Lai

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It should be \[1^3=1^2=1\] in first row to make it (look like) a pattern. \(\ddot\smile\)

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Indeed it should, thanks for spotting that!

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General term

\[ (n^2 - n + 1) + (n^2 - n + 3) + ..... + (n^2 + n -1) = n^3\]

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Can you give the example please ??

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what are you not able to understand , can you tell?

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\(\sum_{i=1}^{n} \left(n(n-1)-1+2i\right) = n\times n(n-1)-n+n(n+1) = n^3\)

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The cube of an integer \(n\) can be written as sum of \(n\) consecutive odd numbers, first odd number being \((n-1)^2-n\).

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By the way, a good pattern and the thing I observed is There is a difference of 2 between consecutive numbers

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Odd number means difference of 2.

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Can you explain it to me ?? For additional information for me, hehehe...

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