04 Aug '08 12:26

Prove that for any positive integer n

n cubed equals the sum of n sequential odd numbers

eg 4^3 = 64 = 13 + 15 + 17 + 19

n cubed equals the sum of n sequential odd numbers

eg 4^3 = 64 = 13 + 15 + 17 + 19

- Joined
- 09 Jun '07
- Moves
- 46061

at home- Joined
- 15 Feb '07
- Moves
- 667

- Joined
- 29 Apr '05
- Moves
- 827

- Joined
- 15 Feb '07
- Moves
- 667

04 Aug '08 16:06

1^3 = 1 = 1*Originally posted by crazyblue***isnt 3 a positive integer? 3^3 = 27 and i dont see how 4 odd numbers can add up to another odd number. am i missing something, or did you mean any _even_ positive integer?**

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

Pattern continues.

Prove that the sum of the first n cubes is the square of the sum of the first n numbers.- Joined
- 10 Dec '06
- Moves
- 7733

podunk, PA04 Aug '08 16:45

proofing aside,....... because i can't do it๐........That is pretty cool๐ฒ*Originally posted by geepamoogle***1^3 = 1 = 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

Pattern continues.

Prove that the sum of the first n cubes is the square of the sum of the first n numbers.- Joined
- 29 Apr '05
- Moves
- 827

- Joined
- 12 Sep '07
- Moves
- 2668

05 Aug '08 04:411 editOoo geep i love that relation. I especially like how it doesn't generalize to higher powers. Can be proven via induction(boring and ugly), telescoping(could be messy), a nice substitution (pretty cool), and geometrically (very awesome).

Sum(i^3, i, 1, n)=Sum(i, i, 1, n)^2 Would look better if we had LATEX.- Joined
- 11 Nov '05
- Moves
- 43938

05 Aug '08 13:02

My wife doesn't like latex. She's more into ... well, that's personal...*Originally posted by Dejection***Ooo geep i love that relation. I especially like how it doesn't generalize to higher powers. Can be proven via induction(boring and ugly), telescoping(could be messy), a nice substitution (pretty cool), and geometrically (very awesome).**

Sum(i^3, i, 1, n)=Sum(i, i, 1, n)^2 Would look better if we had LATEX.- Joined
- 10 Dec '06
- Moves
- 7733

podunk, PA05 Aug '08 13:481 edithey what i am doing is just to gain a litle more insight..(ie completely unnecessary)..

so when trying to find n^3 I noticed that the last odd term in the series

is

(n^2 + n -1) = A(n)

and from there I deduced the first term of the series as

(n^2 + n -1) -2( n - 1 )

simplifying to

n^2 -n + 1 = A(1)

then use the formula for summing the first "n" terms of an arithmetic sequence

S(n) = n / 2 ( A(1) + A(n) )

= n^3

Like I said....Hooray for me๐....... I know, haha...........just wanted to see if anyone else looked at it the same way?

As I said , this may seem trivial and silly, but for me it wasn't ....SO BACK OFF๐ ....haha, I took that line from Billy Madison- Joined
- 10 Aug '07
- Moves
- 32581

18 Aug '08 11:02Let me start with this equation:

(1) - 1+3+5+...+n = [(n+1)/2]^2

If you take two numbers x, y, with y>x,

y^2 - x^2 will always be a number of consecutive odd numbers.

Now, if I prove

n^3 = y^2 - x^2 for y>x, it is sufficient.

We also know that:

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

I have rewritten this as:

Sum (n^3) = [n(n+1)/2]^2

Let us take m=n-1,

Now,

n^3 = Sum (n^3) - Sum (m^3) = p^2 - q^2

where, p = n(n+1)/2 and q = m(m+1)/2

p > q, and hence for any number n,

n^3 will always be a sum of a number of consecutive odd numbers.

Looks like I got a proof, though I am using two well known formula (1) and (2).

Wolf, Let me know if this is good.

+Udaya- Joined
- 12 Sep '07
- Moves
- 2668

- Joined
- 10 Aug '07
- Moves
- 32581

19 Aug '08 09:49Here is another proof:

n^3 = n * n^2

Sum of an arithmetic series is:

S = n * m, where n = number of terms, and m = mean of the series.

In our case, in the series of consecutive odd numbers,

if m = n^2, it will satisfy.

For the series (of consecutive odd numbers), interval d=2.

Also,

Mean m = [2*a + (n-1)*d]/2 where a = first term, and d=interval.

In our case d = 2, the equation reduces to:

m = a + (n-1).

Now, it is sufficient to prove that 'a' is odd. Then n^3 is the sum of a series of consecutive odd numbers.

If n is odd, m=n^2 is also odd. n-1 is even, so 'a' is odd.

If n is even, m=n^2 is also even. n-1 is odd, so 'a' must be odd to satisfy the equation.

I think, this is better than the first one. No formulae used.

+Udaya- Joined
- 16 Feb '06
- Moves
- 115301

Shanghai19 Aug '08 11:14Imagnine an nxnxn cube made from those 1 unit squares that you use at primary school when learning to count. There are n squares nxn.

If n is odd keep one of these squares, take 2 from one square add it to another. Take 4 from another square add it to another. Take 6 from one square add it to another and so on.

If n is even take 1 from one square and add it to another. Take 3 from another square and add it to another. Take 5 from another and so on.

Both will give you a piles with consecutive odd numbers of unit squares.- Joined
- 12 Sep '07
- Moves
- 2668

- Joined
- 16 Feb '06
- Moves
- 115301

Shanghai19 Aug '08 16:43As an example, a cube 8x8x8. made of little unit cubes. This can be split into 8 layers with 8x8 little cubes. (each has 64).

take one cube from one and put it with another (63,65)

take 3 cubes from one and put with another (61,67)

5 cubes from one to another (59,69)

7 cubes from one to another (57,71)

You get 57,59,61,63,65,67,69,71

Now if it is 7x7x7 cube made of unit cubes

Split into 7 layers each with 7x7 (49 cubes)

Leave one alone (49)

Move 2 from one to another (47,51)

Move 4 from one to another (45,53)

Move 6 from one to another (43,55)

You get 43,45,47,49,51,53,55

It was my attempt to come up with a more kinaesthetic answer but it is a bit tricky without good graphics or unit cubes!