Cubic Number Pattern

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.

03=00^{3} = 0

13=11^{3} = 1

23=8=3+52^{3} = 8 = 3 + 5

33=27=7+9+113^{3} = 27 = 7 + 9 + 11

43=64=13+15+17+194^{3} = 64 = 13 + 15 + 17 + 19

53=125=21+23+25+27+295^{3} = 125 = 21 + 23 + 25 +27 + 29

63=216=31+33+35+37+39+416^{3} = 216 = 31 + 33 + 35 + 37 + 39 + 41

73=343=43+45+47+49+51+53+557^{3} = 343 = 43 + 45 + 47 + 49 +51 + 53 + 55

83=512=57+59+61+63+65+67+69+718^{3} = 512 = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71

93=729=73+75+77+79+81+83+85+87+899^{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...

Note by Jonathan Christianto
4 years, 9 months ago

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1 vote

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Let me give you a pattern:

13=12=11^{3} = 1^{2} = 1

13+23=(1+2)2=91^{3}+2^{3} = (1+2)^{2} = 9

13+23+33=(1+2+3)2=361^{3}+2^{3}+3^{3} = (1+2+3)^{2} = 36

13+23+33+43=(1+2+3+4)2=1001^{3}+2^{3}+3^{3}+4^{3} = (1+2+3+4)^{2} = 100

In general,

13+23++n3=(1+2++n)2\boxed{1^{3}+2^{3}+\cdots+n^{3} = (1+2+\cdots+n)^{2}}

Can you figure out why this is? ;)

Jake Lai - 4 years, 9 months ago

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It should be 13=12=11^3=1^2=1 in first row to make it (look like) a pattern. ¨\ddot\smile

Pranjal Jain - 4 years, 9 months ago

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Indeed it should, thanks for spotting that!

Jake Lai - 4 years, 9 months ago

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In boxed part, 13+231^3+2^3

Pranjal Jain - 4 years, 9 months ago

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@Jake Lai

Pranjal Jain - 4 years, 9 months ago

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I have edited it! ¨\ddot\smile

Pranjal Jain - 4 years, 9 months ago

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

Sudhir Aripirala - 4 years, 9 months ago

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k=0n1(n2n+1+2k)=n3\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 n2{ n }^{ 2 }

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

k=0n1(n3n+1+2k)=n4\displaystyle \sum _{ k=0 }^{ n-1 }{ \left( { n }^{ 3 }-n+1+2k \right) ={ n }^{ 4 } }

In fact, for any aa, this is true

k=0n1(nan+1+2k)=na+1\displaystyle \sum _{ k=0 }^{ n-1 }{ \left( { n }^{ a }-n+1+2k \right) ={ n }^{ a+1 } }

Michael Mendrin - 4 years, 9 months ago

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Correct me if I am wrong. So, for n2n^{2} it will be

k=0n1(n+1+2k)=n2\sum_{k=0}^{n-1} (n+1+2k) = n^{2}

Jonathan Christianto - 4 years, 9 months ago

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Actually, leave out nn, since n1n=0{n}^{1}-n=0, and we have the classic sum of odd numbers 1+2k1+2k to make squares.

Michael Mendrin - 4 years, 9 months ago

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Thanks for formula for ANY power.

Niranjan Khanderia - 4 years, 6 months ago

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General term

(n2n+1)+(n2n+3)+.....+(n2+n1)=n3 (n^2 - n + 1) + (n^2 - n + 3) + ..... + (n^2 + n -1) = n^3

U Z - 4 years, 9 months ago

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Can you give the example please ??

Jonathan Christianto - 4 years, 9 months ago

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

U Z - 4 years, 9 months ago

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@U Z You have given me the general term. Can you give the example ?? For example, the result of 333^{3} using the general term. Can you catch it ??

Jonathan Christianto - 4 years, 9 months ago

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@Jonathan Christianto 33=(323+1)+(323+3)+(323+5)=7+9+11=273^3=(3^2-3+1)+(3^2-3+3)+(3^2-3+5)\\=7+9+11\\=27

Pranjal Jain - 4 years, 9 months ago

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@Pranjal Jain I got it now, thanks...

Jonathan Christianto - 4 years, 9 months ago

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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

Sudhir Aripirala - 4 years, 9 months ago

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

Jonathan Christianto - 4 years, 9 months ago

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Odd number means difference of 2.

Niranjan Khanderia - 4 years, 6 months ago

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The cube of an integer nn can be written as sum of nn consecutive odd numbers, first odd number being (n1)2n(n-1)^2-n.

Pranjal Jain - 4 years, 9 months ago

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i=1n(n(n1)1+2i)=n×n(n1)n+n(n+1)=n3\sum_{i=1}^{n} \left(n(n-1)-1+2i\right) = n\times n(n-1)-n+n(n+1) = n^3

Janardhanan Sivaramakrishnan - 4 years, 9 months ago

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