 Posers and Puzzles

1. 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
2. 04 Aug '08 13:25
Finding the two following equations works wonders..

SUM(i=1 to n, i)
SUM(i=1 to n, i, i^3)
3. 04 Aug '08 15:55
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?
4. 04 Aug '08 16:06
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?
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.
5. 04 Aug '08 16:45
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.
proofing aside,....... because i can't do it😀........That is pretty cool😲
6. 04 Aug '08 22:21
oh nevermind, i had a blind moment up there 😉
7. 05 Aug '08 04:411 edit
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.
8. 05 Aug '08 13:02
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.
My wife doesn't like latex. She's more into ... well, that's personal...
9. 05 Aug '08 13:481 edit
hey 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 me😏....... 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 OFF😠....haha, I took that line from Billy Madison
10. 18 Aug '08 11:02

(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
11. 18 Aug '08 12:22
Looks good to me.
12. 19 Aug '08 09:49
Here 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
13. 19 Aug '08 11:14
Imagnine 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.
14. 19 Aug '08 13:20
I didn't get that, could you explain please?
15. 19 Aug '08 16:43
As 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!