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# 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.

$$0^{3} = 0$$

$$1^{3} = 1$$

$$2^{3} = 8 = 3 + 5$$

$$3^{3} = 27 = 7 + 9 + 11$$

$$4^{3} = 64 = 13 + 15 + 17 + 19$$

$$5^{3} = 125 = 21 + 23 + 25 +27 + 29$$

$$6^{3} = 216 = 31 + 33 + 35 + 37 + 39 + 41$$

$$7^{3} = 343 = 43 + 45 + 47 + 49 +51 + 53 + 55$$

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

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

and so on and on and on !

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

Note by Jonathan Christianto
2 years, 3 months ago

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

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

$$\displaystyle \sum _{ k=0 }^{ n-1 }{ \left( { n }^{ a }-n+1+2k \right) ={ n }^{ a+1 } }$$ · 2 years, 3 months ago

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

$$\sum_{k=0}^{n-1} (n+1+2k) = n^{2}$$ · 2 years, 3 months ago

Actually, leave out $$n$$, since $${n}^{1}-n=0$$, and we have the classic sum of odd numbers $$1+2k$$ to make squares. · 2 years, 3 months ago

Thanks for formula for ANY power. · 2 years ago

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? ;) · 2 years, 3 months ago

Sum of cubes = n(n+1)(n)(n+1)/4= [n(n+1)/2]^2= [Sum of the numbers]^2 · 2 years, 3 months ago

In boxed part, $$1^3+2^3$$ · 2 years, 3 months ago

I have edited it! $$\ddot\smile$$ · 2 years, 3 months ago

@Jake Lai · 2 years, 3 months ago

It should be $1^3=1^2=1$ in first row to make it (look like) a pattern. $$\ddot\smile$$ · 2 years, 3 months ago

Indeed it should, thanks for spotting that! · 2 years, 3 months ago

General term

$(n^2 - n + 1) + (n^2 - n + 3) + ..... + (n^2 + n -1) = n^3$ · 2 years, 3 months ago

Can you give the example please ?? · 2 years, 3 months ago

what are you not able to understand , can you tell? · 2 years, 3 months ago

You have given me the general term. Can you give the example ?? For example, the result of $3^{3}$ using the general term. Can you catch it ?? · 2 years, 3 months ago

$$3^3=(3^2-3+1)+(3^2-3+3)+(3^2-3+5)\\=7+9+11\\=27$$ · 2 years, 3 months ago

I got it now, thanks... · 2 years, 3 months ago

$$\sum_{i=1}^{n} \left(n(n-1)-1+2i\right) = n\times n(n-1)-n+n(n+1) = n^3$$ · 2 years, 3 months ago

The cube of an integer $$n$$ can be written as sum of $$n$$ consecutive odd numbers, first odd number being $$(n-1)^2-n$$. · 2 years, 3 months ago

By the way, a good pattern and the thing I observed is There is a difference of 2 between consecutive numbers · 2 years, 3 months ago

Odd number means difference of 2. · 2 years ago