Posers and Puzzles

Posers and Puzzles

  1. Standard memberwolfgang59
    Mr. Wolf
    at home
    Joined
    09 Jun '07
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    46061
    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. Joined
    15 Feb '07
    Moves
    667
    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. Joined
    29 Apr '05
    Moves
    827
    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. Joined
    15 Feb '07
    Moves
    667
    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. Subscriberjoe shmo
    Strange Egg
    podunk, PA
    Joined
    10 Dec '06
    Moves
    7733
    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 it๐Ÿ˜€........That is pretty cool๐Ÿ˜ฒ
  6. Joined
    29 Apr '05
    Moves
    827
    04 Aug '08 22:21
    oh nevermind, i had a blind moment up there ๐Ÿ˜‰
  7. Joined
    12 Sep '07
    Moves
    2668
    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. Joined
    11 Nov '05
    Moves
    43938
    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. Subscriberjoe shmo
    Strange Egg
    podunk, PA
    Joined
    10 Dec '06
    Moves
    7733
    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 me๐Ÿ˜....... 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 OFF๐Ÿ˜ ....haha, I took that line from Billy Madison
  10. Joined
    10 Aug '07
    Moves
    32581
    18 Aug '08 11:02
    Let 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
  11. Joined
    12 Sep '07
    Moves
    2668
    18 Aug '08 12:22
    Looks good to me.
  12. Joined
    10 Aug '07
    Moves
    32581
    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. Shanghai
    Joined
    16 Feb '06
    Moves
    115301
    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. Joined
    12 Sep '07
    Moves
    2668
    19 Aug '08 13:20
    I didn't get that, could you explain please?
  15. Shanghai
    Joined
    16 Feb '06
    Moves
    115301
    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!
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