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Math Help - Geometric progression

  1. #1
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    Geometric progression

    Just a quick question;

    How do I show that (1+2+...+n)^2 = 1^3+2^3+...+n^3 ?
    Last edited by weasley74; March 25th 2008 at 01:48 PM.
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    Quote Originally Posted by weasley74 View Post
    Just a quick question;

    How do I show that (1+2+...+n)^2 = 1^3+2^3+...+n^3 ?
    The sum on the right can be shoen to be a quartic in n (construct the
    difference table and it becomes obviouse).

    Which quartic can be established by fitting the first few terms to the
    quartic.

    The sum on the left is: \left(\frac{n(n+1)}{2}\right)^2.

    Expand and show the required equality.

    RonL
    Last edited by CaptainBlack; March 26th 2008 at 04:29 AM.
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    Quote Originally Posted by weasley74 View Post
    Just a quick question;

    How do I show that (1+2+...+n)^2 = 1^3+2^3+...+n^3 ?
    Second method from the department of incredibly elegant but indirect proofs.

    The expression on the left is obviously a quartic in n, since the table of
    second differences of what is inside the square is constant it is a quadratic in
    n and so its square is a quartic in n.

    As explained before what is on the right is also a quartic.

    To show that two quartics are identical you need only show that they are
    equal at five distinct points. So compute the first 5 values of what is on the
    left and the corresponding values for what is on the right and if the
    corresponding terms are equal the identity is proven.

    RonL
    Last edited by CaptainBlack; March 26th 2008 at 05:59 AM. Reason: correcting spelling and typos
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    Quote Originally Posted by CaptainBlack View Post
    Second method from the department of incredibly elegant but indirect proofs.

    The expression on the left is obviously a quatic in n, since the table of
    second differences of what is inside the square is constant it is a quadratic in
    n and so its square is a quartic in n.

    As explained before what is on the right is also a quartic.

    To show that two quartics are identical you need only show that they are
    equal at five distinct points. So comopute the first 5 values of what is on the
    left and the corresponding values for what is on the right and if the
    corresponding terms are equal the identity is proven.

    RonL
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    So I just have to show that the five first values are equal? it sounds a bit too easy..
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    Quote Originally Posted by weasley74 View Post
    So I just have to show that the five first values are equal? it sounds a bit too easy..
    Well, if you want something more "logic" oriented, you could always do a proof using Mathematical Induction.

    -Dan
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    Quote Originally Posted by topsquark View Post
    Well, if you want something more "logic" oriented, you could always do a proof using Mathematical Induction.

    -Dan
    and how do I do that? =)
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    Quote Originally Posted by topsquark View Post
    Well, if you want something more "logic" oriented, you could always do a proof using Mathematical Induction.

    -Dan

    I was just about to post method 3, Mathematical Induction (which I can
    confirm does work quite nicely, but is not as elegant as method 2).

    RonL
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    Quote Originally Posted by weasley74 View Post
    Just a quick question;

    How do I show that (1+2+...+n)^2 = 1^3+2^3+...+n^3 ?
    Let n = 1.
    (1)^2 = 1^3? True.

    So assume the theorem is true for some n = k. Then we need to prove that it is true for n = k + 1.

    Our assumption is that
    (1 + 2 + ~...~ + k)^2 = 1^3 + 2^3 + ~...~ + k^3

    What we wish to prove is
    (1 + 2 + ~...~ + k + (k + 1))^2 = 1^3 + 2^3 +~...~+k^3 + (k + 1)^3

    Now,
    (1 + 2 + ~...~ + k + (k + 1))^2 = (1 + 2 + ~...~ + k)^2 + (k + 1)^2 + 2(1 \cdot (k + 1) + 2 \cdot (k + 1) + ~...~ k(k + 1))

    = (1^3 + 2^3 + ~...~ + k^3) + (k + 1)^2 + 2(1 \cdot (k + 1) + 2 \cdot (k + 1) + ~...~ k(k + 1))
    according to our assumption. So plugging this back into what we need to prove, we see that we need to show
    = (1^3 + 2^3 + ~...~ + k^3) + (k + 1)^2 + 2(1 \cdot (k + 1) + 2 \cdot (k + 1) + ~...~ k(k + 1))  = 1^3 + 2^3 +~...~+k^3 + (k + 1)^3

    or
    (k + 1)^2 + 2(1 \cdot (k + 1) + 2 \cdot (k + 1) + ~...~ k(k + 1)) = (k + 1)^3

    So let's work a bit more on that left hand side.
    (k + 1)^2 + 2(1 \cdot (k + 1) + 2 \cdot (k + 1) + ~...~ k(k + 1))

    = (k^2 + 2k + 1) + 2(k + 1)(1 + 2 + ~...~ + k)

    and the sum of the first k numbers is
    = (k^2 + 2k + 1) + 2(k + 1) \frac{k(k + 1)}{2}

    = (k^2 + 2k + 1) + k(k + 1)^2

    = k^2 + 2k + 1 + k^3 + 2k^2 + k

    = k^3 + 3k^2 + 3k + 1

    = (k + 1)^3
    as we needed.

    So the theorem is true for n = 1 and thus for all integers by induction.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    I was just about to post method 3, Mathematical Induction (which I can
    confirm does work quite nicely, but is not as elegant as method 2).

    RonL
    Agreed. And Mathematical Induction here is a bit tedious.

    -Dan
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    Quote Originally Posted by weasley74 View Post
    and how do I do that? =)
    First you should know how induction works:

    You have a proposition about a netural number n, such as your identity.

    Show it holds for a base case, typical n=1 or n=0.

    The if you assume that it hold for some k, show this implies it holds for k+1

    Then you are done the general proposition is proven for all n greater than or
    equal to the base case.

    Checking the base case is trivial in this case as for n=1 both sides of the
    equality are 1.

    I will let you have a go at proving the induction step its not difficult.

    RonL
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    Quote Originally Posted by weasley74 View Post
    So I just have to show that the five first values are equal? it sounds a bit too easy..
    That part is easy, the key part of the argument is identifying that both
    sides are quartics.

    RonL
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    Quote Originally Posted by topsquark View Post
    Agreed. And Mathematical Induction here is a bit tedious.

    -Dan
    On my a5 note pad its half a side of paper!

    RonL
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