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Math Help - Interesting equality

  1. #1
    TD!
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    Interesting equality

    I remember that quite some time ago, when I first saw this, I didn't believe it was true. Of course, trying for a few values made me see that it was probably true, so all that was left was proving it - now that's up to you

    Prove that for all natural n, the following holds:

    \left( {\sum\limits_{i = 1}^n i } \right)^2  = \sum\limits_{i = 1}^n {i^3 }

    For those who don't like summation symbols, the above is just math language to say that:

    \left( {1 + 2 + ... + n} \right)^2  = 1^3  + 2^3  + ... + n^3

    Please don't reveal your answer yet, or post it 'covered'.
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    Grand Panjandrum
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    Quote Originally Posted by TD!
    I remember that quite some time ago, when I first saw this, I didn't believe it was true. Of course, trying for a few values made me see that it was probably true, so all that was left was proving it - now that's up to you

    Prove that for all natural n, the following holds:

    \left( {\sum\limits_{i = 1}^n i } \right)^2  = \sum\limits_{i = 1}^n {i^3 }

    For those who don't like summation symbols, the above is just math language to say that:

    \left( {1 + 2 + ... + n} \right)^2  = 1^3  + 2^3  + ... + n^3

    Please don't reveal your answer yet, or post it 'covered'.
    and the continuous analogue:

    \left(\int_0^x u\ du \right)^2=\int_0^x u^3\ du,

    which is easier to prove as we are more familiar with the rules for
    manipulating integrals than we are in handling summations.

    RonL
    Last edited by CaptainBlack; January 13th 2006 at 01:29 PM.
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  3. #3
    TD!
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    Indeed, but that one is rather easy to show
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    Grand Panjandrum
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    Quote Originally Posted by TD!
    Indeed, but that one is rather easy to show
    but so is the other if you know the formula for the sum of cubes

    RonL
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  5. #5
    TD!
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    Quote Originally Posted by CaptainBlack
    but so is the other if you know the formula for the sum of cubes

    RonL
    Well actually, that's what we're indirectly trying to prove so unfortunately, it's not allowed to use that formula
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    I admit that is an interesting equality, but it is rather easy to proof.

    I have a different question find all integral solutions for x,y
    such as,
    \sum^n_{k=1}k^x= (\sum^n_{k=1}k)^y
    for all n\in Z^+

    For example (x,y)=(3,2)
    But are there more?
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by TD! View Post
    I remember that quite some time ago, when I first saw this, I didn't believe it was true. Of course, trying for a few values made me see that it was probably true, so all that was left was proving it - now that's up to you

    Prove that for all natural n, the following holds:

    \left( {\sum\limits_{i = 1}^n i } \right)^2  = \sum\limits_{i = 1}^n {i^3 }

    For those who don't like summation symbols, the above is just math language to say that:

    \left( {1 + 2 + ... + n} \right)^2  = 1^3  + 2^3  + ... + n^3

    Please don't reveal your answer yet, or post it 'covered'.
    we could use induction on the following:
    \left( {1 + 2 + ... + n} \right)^2  = 1^3  + 2^3  + ... + n^3
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    Quote Originally Posted by Jhevon View Post
    we could use induction on the following:
    \left( {1 + 2 + ... + n} \right)^2  = 1^3  + 2^3  + ... + n^3
    Or you can use addition formulas.

    \sum_{k=1}^n k = \frac{n(n+1)}{2}

    \sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4}.

    I seen this identity like 15 times throughout my life.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Or you can use addition formulas.

    \sum_{k=1}^n k = \frac{n(n+1)}{2}

    \sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4}.

    I seen this identity like 15 times throughout my life.
    That's true, but i believe TD! said in an earlier post that we weren't allowed to use the formulas

    Quote Originally Posted by TD! View Post
    Well actually, that's what we're indirectly trying to prove so unfortunately, it's not allowed to use that formula
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  10. #10
    TD!
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    Induction would be a good approach, yes.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by TD! View Post
    Induction would be a good approach, yes.
    were you thinking of something else?
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    How about these identities?


    2\left(\sum^n_1 k\right)^4\;=\;\sum^n_1 k^5 + \sum^n_1 k^7

    2(1 + 2 + 3 + \cdots + n)^4 \;=\;\left(1^5 + 2^5 + 3^5 + \hdots + n^5\right) + \left(1^7 + 2^7 + 3^7 + \hdots + n^7\right)

    The sum of the first n positive integers, raised to the fourth power,
    . . is the mean of the sum of the 5th powers and the sum of the 7th powers.



    Let T_n be the n^{th} triangular number: . T_n = \frac{n(n+1)}{2}

    3\left[\sum^n_1 T_k\right]^3 \;=\;\sum^n_1\left(T_k\right)^3 + 2\!\cdot\!\sum^n_1\left(T_k\right)^4

    3(1\ +\ 3\ +\ 6\ +\ \cdots\ +\ T_n)^3 \;=\;\left(1^3 + 3^3 + 6^3 + \cdots + T_n^3 \right)\,+\,2\left(1^4 + 3^4 + 6^4 + \cdots + T_n^4 \right)

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  13. #13
    TD!
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    Quote Originally Posted by Jhevon View Post
    were you thinking of something else?
    No, induction was what I had in mind and what I used.
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  14. #14
    Super Member PaulRS's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I admit that is an interesting equality, but it is rather easy to proof.

    I have a different question find all integral solutions for x,y
    such as,
    \sum^n_{k=1}k^x= (\sum^n_{k=1}k)^y
    for all n\in Z^+

    For example (x,y)=(3,2)
    But are there more?
    No

    \sum_{k=1}^n{k^r}\sim{\frac{n^{r+1}}{r+1}} for r\geq{0} (asymptotically equal)

    <br />
{\text{As a necessary condition we have:}}\left( {\sum\limits_{k = 1}^n k } \right)^y  \sim \sum\limits_{k = 1}^n {k^x } <br /> <br />

    Since <br />
\left( {\sum\limits_{k = 1}^n k } \right)^y  \sim \frac{{n^{2y} }}<br />
{{2^y }},\sum\limits_{k = 1}^n {k^x }  \sim \frac{{n^{x + 1} }}<br />
{{2^{x + 1} }}<br />

    It is equivalent to <br />
\left( {\sum\limits_{k = 1}^n k } \right)^y  \sim \sum\limits_{k = 1}^n {k^x }  \Leftrightarrow \frac{{n^{2y} }}<br />
{{2^y }} \sim \frac{{n^{x + 1} }}<br />
{{x + 1}}

    So it must be: <br />
2y = x + 1{\text{ and }}2^y  = x + 1<br />

    <br />
\begin{gathered}<br />
  {\text{It is easy to verify* that }}2^{y - 1}  = y{\text{ has only 2 roots}} \hfill \\<br />
  y = 1,y = 2 \hfill \\<br />
  {\text{Thus }}\left( {x,y} \right) = (1,1){\text{ and }}\left( {x,y} \right) = (3,2){\text{ are the only}} \hfill \\<br />
  {\text{solutions}} \hfill \\ <br />
\end{gathered} <br />

    * Derivatives



    THe original problem
    <br />
\left( {\sum\limits_{k = 1}^n k } \right)^2  = \sum\limits_{k = 1}^n {k^2 }  + 2 \cdot \sum\limits_{1 \leqslant k < j \leqslant n} {\left( {k \cdot j} \right)} <br />
    <br />
\sum\limits_{1 \leqslant k < j \leqslant n} {\left( {k \cdot j} \right)}  = \sum\limits_{k = 1}^n {\left( {k \cdot \sum\limits_{j =0}^{k - 1} j } \right)}  = \sum\limits_{k = 1}^n {\left( {k \cdot \tfrac{{k \cdot \left( {k - 1} \right)}}<br />
{2}} \right)} <br />
    <br />
\sum\limits_{1 \leqslant k < j \leqslant n} {\left( {k \cdot j} \right)}  = \tfrac{1}<br />
{2} \cdot \sum\limits_{k = 1}^n {\left( {k^3  - k^2 } \right)}  = \tfrac{1}<br />
{2} \cdot \sum\limits_{k = 1}^n {\left( {k^3 } \right)}  - \tfrac{1}<br />
{2} \cdot \sum\limits_{k = 1}^n {\left( {k^2 } \right)} <br />
    <br />
\left( {\sum\limits_{k = 1}^n k } \right)^2  = \sum\limits_{k = 1}^n {k^2 }  + 2 \cdot \left[ {\tfrac{1}<br />
{2} \cdot \sum\limits_{k = 1}^n {\left( {k^3 } \right)}  - \tfrac{1}<br />
{2} \cdot \sum\limits_{k = 1}^n {\left( {k^2 } \right)} } \right] = \sum\limits_{k = 1}^n {\left( {k^3 } \right)} \blacksquare <br />

    Actually I used the identity \sum_{k=1}^n{k^2}=\frac{n\cdot{(n+1)}\cdot{(2n+1)}  }{6} once in a math competition, but this one \sum_{k=1}^n{k}=\frac{n\cdot{(n+1)}}{2} comes round all the time
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