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Math Help - proving the summation of a finite series .

  1. #1
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    proving the summation of a finite series .

    FInd \sum^{n}_{r=1}r^3

    My attempt :

    Using this identity : r^4-(r-1)^4=4r^3-6r^2+4r-1

    \sum^{n}_{r=1}r^4-(r-1)^4=\sum^{n}_{r=1}4r^3-6r^2+4r-1

    The LHS can be written as n^4 .

     4\sum^{n}_{r=1}r^3=n^4+n(n+1)(2n+1)+2n(n+1)-n

    but it is not anywhere nearer to \frac{1}{4}n^2(n+1)^2 ,there must be some problem with my proving .
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  2. #2
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    Quote Originally Posted by mathaddict View Post
    FInd \sum^{n}_{r=1}r^3

    My attempt :

    Using this identity : r^4-(r-1)^4=4r^3-6r^2+4r-1

    \sum^{n}_{r=1}r^4-(r-1)^4=\sum^{n}_{r=1}4r^3-6r^2+4r-1

    The LHS can be written as n^4 .

     4\sum^{n}_{r=1}r^3=n^4+n(n+1)(2n+1)+2n(n+1)-n

    but it is not anywhere nearer to \frac{1}{4}n^2(n+1)^2 ,there must be some problem with my proving .
    You have made a couple of sign errors. The last line should be:

     4\sum^{n}_{r=1}r^3 = n^4+n(n+1)(2n+1) {\color{red} - } 2n(n+1) {\color{red} + } n

    and the right hand side obligingly factorises into n^2 (n+1)^2.
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  3. #3
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    Thanks Mr F , i try to factorise and i still can't get the proof right .

     <br />
4\sum^{n}_{r=1}r^3 = n^4+n(n+1)(2n+1)-2n(n+1)+n<br />
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  4. #4
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    Hello, mathaddict!

    I'll modify your work ... and include all the steps.


    FInd: . \sum^{n}_{r=1}r^3

    Using this identity: . r^4-(r-1)^4\:=\:4r^3-6r^2+4r-1

    . . we have: . \sum^{n}_{r=1}\bigg[r^4-(r-1)^4\bigg]\;=\;\sum^{n}_{r=1}\bigg[4r^3-6r^2+4r-1\bigg]

    The LHS can be written as n^4 .


    We have: . n^4 \;=\;4\sum r^3 - 6\sum r^2 + 4\sum r - \sum 1

    . . . . . . . n^4 \;=\;4\sum r^3 - 6\!\cdot\!\tfrac{n(n+1)(2n+1)}{6} + 4\!\cdot\!\tfrac{n(n+1)}{2} - n

    . . . . . . . n^4 \;=\;4\sum r^3 - n(n+1)(2n+1) + 2n(n+1) - n


    \text{We have: }\;4\sum r^3 \;=\;\underbrace{n^4 + n}_{\quad\searrow} + n(n+1)(2n+1) - 2n(n+1)

    . . . . . . 4\sum r^3 \;=\;\overbrace{n(n+1)(n^2-n+1)} + n(n+1)(2n+1) - 2n(n+1)


    Factor: . . 4\sum r^3 \;=\;n(n+1)\bigg[(n^2-n+1) + (2n+1) - 2\bigg]

    . . . . . . . 4\sum r^3 \;=\;n(n+1)\bigg[n^2+n\bigg] \;=\;n(n+1)\cdot n(n+1) \;=\;n^2(n+1)^2


    Therefore: . \sum^n_{r=1}r^3 \;=\;\frac{1}{4}n^2(n+1)^2

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  5. #5
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    Thanks a lot Soroban , btw

    n^4+n=n(n+1)(n^2-n+1) , is it an algebraic identity ?
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  6. #6
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    Yes. Use the fact that: {\color{red}a^3 + b^3} = {\color{blue}(a+b)(a^2 - ab + b^2)}

    So:  n^4 + n \ = \ n({\color{red}n^3 + 1}) \ = \ n {\color{blue}(n+1)(n^2 - n + 1)}
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