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Math Help - (FP1) summation of finite series using standard results

  1. #1
    Member ssadi's Avatar
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    Question (FP1) summation of finite series using standard results

    Find the sum of the series:
     1^2-2^2+3^2-4^2+...-(2n)^2
    I do the sum and each time I get a different answers, none of them matching with the answer in my book.
    Book's answer:-n(2n+1)
    My latest answer: -3n(2n-1)
    My latest working:
    (a)  1^2, 3^2, 5^2, ...(2n-1)^2, so the series can be represented by (2r+1)^2
    starting from r=1, there is (n-1) terms and 1^2 have to be added separately.
    (b) 2^2,4^2,6^2,...(2n)^2, so the series can be represented by (2r)^2
    starting from r=1, there is n terms.
    ----
    (a)  1+\sum\limits_{r=1}^{n-1} (2n-1)^2
    =1+4\sum\limits_{r=1}^{n-1} r^2-4\sum\limits_{r=1}^{n-1} r+\sum\limits_{r=1}^{n-1}1
    =4/3 n^3-4n^2+11/3 n
    (b)  \sum\limits_{r=1}^{n} (2r)^2<br />
=4/3n^3-2n^2+2/3n
    Summation=(a)-(b)=-6n^2+3n
    =-3n(2n-1)
    Can you please narrow down the field where I should search the error?
    I am exasperated looking through my rough handwriting.
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  2. #2
    Moo
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    Hello,

    1^2-2^2+3^2-\dots+(2n-1)^2-(2n)^2

    Quote Originally Posted by ssadi View Post
    (a)  {\color{blue}1}+\sum\limits_{r=1}^{n-1} (2{\color{red}r}-1)^2
    Small modification in red =)
    Question in blue : why did you add 1 ???
    Other question : it doesn't go 'til n-1, but n :
    [2(n-1)-1]^2=[2n-3]^2 \neq (2n-1)^2

    Try to rethink your series with these remarks

    And I guess it's better to leave the summations in their factored form, that is \sum_{k=1}^n ~ k^2=\frac{n(n+1)(2n+1)}{6}

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

    \sum_{k=1}^n ~ 1=n
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  3. #3
    Member ssadi's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    1^2-2^2+3^2-\dots+(2n-1)^2-(2n)^2


    Small modification in red =)
    Question in blue : why did you add 1 ???
    Other question : it doesn't go 'til n-1, but n :
    [2(n-1)-1]^2=[2n-3]^2 \neq (2n-1)^2

    Try to rethink your series with these remarks

    And I guess it's better to leave the summations in their factored form, that is \sum_{k=1}^n ~ k^2=\frac{n(n+1)(2n+1)}{6}

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

    \sum_{k=1}^n ~ 1=n
    I added 1 because when r=1, the series starts from 3^2, and that extra one means one term less in the summation
    Am I right?
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  4. #4
    Moo
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    Quote Originally Posted by ssadi View Post
    I added 1 because when r=1, the series starts from 3^2, and that extra one means one term less in the summation
    Am I right?
    What is 2r-1 if r=1 ?
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  5. #5
    Member ssadi's Avatar
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    Thumbs up

    Quote Originally Posted by Moo View Post
    What is 2r-1 if r=1 ?
    Stop winking and tell me whether I have a mistake in my working.
    (I get it, but my choice was also logical, why would it give a wrong result?
    Pinpoint the error while I do the sum your way, ok?)
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  6. #6
    Moo
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    Quote Originally Posted by ssadi View Post
    Stop winking and tell me whether I have a mistake in my working.
    (I get it, but my choice was also logical, why would it give a wrong result?
    Pinpoint the error while I do the sum your way, ok?)
    Uuuh ?
    Winking was a kind way of asking it ! So that I don't look like a severe person
    There's no problem, I'll point out your mistakes !

    Although your choice was logical, it was wrong, because you were repeating some terms while you were forgetting some others :s
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  7. #7
    Member ssadi's Avatar
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    Quote Originally Posted by Moo View Post
    Uuuh ?
    Winking was a kind way of asking it ! So that I don't look like a severe person
    There's no problem, I'll point out your mistakes !

    Although your choice was logical, it was wrong, because you were repeating some terms while you were forgetting some others :s
    The winking comment wasn't a rebuke
    I throw up my hands in the air, what was I repeating?
    Though I admit I was ridiculously convoluted.
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  8. #8
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    Okay, I'll try to explain !

    You first wrote 1+\sum_{r=1}^{n-1} (2r-1)^2
    But this equals :

    1+1^2+3^2+\dots+(2(n-2)-1)^2+(2(n-1)-1)^2={\color{red}1}+{\color{red}1^2}+3^2+\dots+(2n-5)^2+(2n-3)^2

    This is supposed to equal 1^2+3^2+\dots+(2n-1)^2
    So the red part is repeating !

    Moreover, the last term isn't equal to 2n-1 but to 2n-3. So in fact, the correct formula is :

    1^2+3^2+\dots+(2n-1)^2=\sum_{r=1}^{{\color{red}n}} (2r-1)^2


    Similarly, 2^2+4^2+\dots+(2n)^2=\sum_{r=1}^{n} (2r)^2=4 \sum_{r=1}^n r^2

    ____________________________________
    The sum S you're looking for is :

    \begin{aligned} S &=\sum_{r=1}^{n} (2r-1)^2-4 \sum_{r=1}^n r^2 \\<br />
&={\color{blue}4 \sum_{r=1}^{n} r^2}-4 \sum_{r=1}^{n} r+\sum_{r=1}^{n} 1-{\color{blue}4 \sum_{r=1}^n r^2} \end{aligned}

    The blue stuff simplifies to 0.

    So \begin{aligned} S &=-4 \sum_{r=1}^{n} r+\sum_{r=1}^{n} 1 \\<br />
&=-4 \cdot \frac{n(n+1)}{2}+n \\<br />
&=-2n(n+1)+n \\<br />
&=-n(2n+1) \end{aligned}

    Is it clearer ?
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  9. #9
    Member ssadi's Avatar
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    Quote Originally Posted by Moo View Post
    Okay, I'll try to explain !

    You first wrote 1+\sum_{r=1}^{n-1} (2r-1)^2
    But this equals :

    1+1^2+3^2+\dots+(2(n-2)-1)^2+(2(n-1)-1)^2={\color{red}1}+{\color{red}1^2}+3^2+\dots+(2n-5)^2+(2n-3)^2

    This is supposed to equal 1^2+3^2+\dots+(2n-1)^2
    So the red part is repeating !

    Moreover, the last term isn't equal to 2n-1 but to 2n-3. So in fact, the correct formula is :

    1^2+3^2+\dots+(2n-1)^2=\sum_{r=1}^{{\color{red}n}} (2r-1)^2


    Similarly, 2^2+4^2+\dots+(2n)^2=\sum_{r=1}^{n} (2r)^2=4 \sum_{r=1}^n r^2

    ____________________________________
    The sum S you're looking for is :

    \begin{aligned} S &=\sum_{r=1}^{n} (2r-1)^2-4 \sum_{r=1}^n r^2 \\<br />
&={\color{blue}4 \sum_{r=1}^{n} r^2}-4 \sum_{r=1}^{n} r+\sum_{r=1}^{n} 1-{\color{blue}4 \sum_{r=1}^n r^2} \end{aligned}

    The blue stuff simplifies to 0.

    So \begin{aligned} S &=-4 \sum_{r=1}^{n} r+\sum_{r=1}^{n} 1 \\<br />
&=-4 \cdot \frac{n(n+1)}{2}+n \\<br />
&=-2n(n+1)+n \\<br />
&=-n(2n+1) \end{aligned}

    Is it clearer ?
    I wrote
    <br /> <br /> <br />

    " alt="1+\sum_{r=1}^{n-1} (2r-1)^2
    " /> here but did the sum first time with this


    <br /> <br /> <br />
    " alt="1+\sum_{r=1}^{n-1} (2r+1)^2
    " />
    So I retained r=1 in 2r+1=3 and the series goes when I switched to (2r-1)^2 second time when I did the sum.

    My head is feeling light, I wonder why

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  10. #10
    Member ssadi's Avatar
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    Quote Originally Posted by Moo View Post
    Okay, I'll try to explain !

    You first wrote 1+\sum_{r=1}^{n-1} (2r-1)^2
    But this equals :

    1+1^2+3^2+\dots+(2(n-2)-1)^2+(2(n-1)-1)^2={\color{red}1}+{\color{red}1^2}+3^2+\dots+(2n-5)^2+(2n-3)^2

    This is supposed to equal 1^2+3^2+\dots+(2n-1)^2
    So the red part is repeating !

    Moreover, the last term isn't equal to 2n-1 but to 2n-3. So in fact, the correct formula is :

    1^2+3^2+\dots+(2n-1)^2=\sum_{r=1}^{{\color{red}n}} (2r-1)^2


    Similarly, 2^2+4^2+\dots+(2n)^2=\sum_{r=1}^{n} (2r)^2=4 \sum_{r=1}^n r^2

    ____________________________________
    The sum S you're looking for is :

    \begin{aligned} S &=\sum_{r=1}^{n} (2r-1)^2-4 \sum_{r=1}^n r^2 \\<br />
&={\color{blue}4 \sum_{r=1}^{n} r^2}-4 \sum_{r=1}^{n} r+\sum_{r=1}^{n} 1-{\color{blue}4 \sum_{r=1}^n r^2} \end{aligned}

    The blue stuff simplifies to 0.

    So \begin{aligned} S &=-4 \sum_{r=1}^{n} r+\sum_{r=1}^{n} 1 \\<br />
&=-4 \cdot \frac{n(n+1)}{2}+n \\<br />
&=-2n(n+1)+n \\<br />
&=-n(2n+1) \end{aligned}

    Is it clearer ?
    I wrote
    <br /> <br /> <br />
1+\sum_{r=1}^{n-1} (2r-1)^2<br />
here but did the sum first time with this


    <br /> <br /> <br />
1+\sum_{r=1}^{n-1} (2r+1)^2<br />
    So I retained r=1 in 2r+1=3 and the series goes when I switched to (2r-1)^2 second time when I did the sum.

    My head is feeling light, I wonder why

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