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Math Help - Sum of series

  1. #16
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    Quote Originally Posted by Moo View Post
    Hello



    n^2+9n+20=(n+4)(n+5)

    \frac{-1}{(n+4)(n+5)}=\frac{n-n+4-5}{(n+4)(n+5)}=\frac{(n+4)-(n+5)}{(n+4)(n+5)}
    would you mind explaining that again. I dont get how you get that?

    Thanks
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  2. #17
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    Following up on TheEmptySet's post.

    We have \sum\limits_{k\,=\,1}^{n}{\left( a_{k+1}-a_{k} \right)}=\sum\limits_{k\,=\,1}^{n}{a_{k+1}}-\sum\limits_{k\,=\,1}^{n}{a_{k}}.

    Since

    \begin{aligned}<br />
   \sum\limits_{k\,=\,1}^{n}{a_{k+1}}&=\sum\limits_{k  \,=\,2}^{n+1}{a_{k}} \\ <br />
 & =\left( \sum\limits_{k\,=\,2}^{n}{a_{k}} \right)+a_{n+1} \\ <br />
 & =\left( \sum\limits_{k\,=\,1}^{n}{a_{k}} \right)-a_{1}+a_{n+1},<br />
\end{aligned}

    back substitute and \sum\limits_{k\,=\,1}^{n}{\left( a_{k+1}-a_{k} \right)}=\left( \sum\limits_{k\,=\,1}^{n}{a_{k}} \right)-a_{1}+a_{n+1}-\left( \sum\limits_{k\,=\,1}^{n}{a_{k}} \right)=a_{n+1}-a_{1}.

    This result is called as "telescoping sum."
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  3. #18
    Moo
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    Quote Originally Posted by taurus View Post
    would you mind explaining that again. I dont get how you get that?

    Thanks
    Do you agree that n^2+9n+20=(n+4)(n+5) ?



    Now, let's find out A and B such that :

    \frac{-1}{(n+4)(n+5)}=\frac{A}{n+4}+\frac{B}{n+5}

    \frac{A}{n+4}+\frac{B}{n+5}=\frac{A(n+5)+B(n+4)}{(  n+4)(n+5)}=\frac{An+5A+Bn+4A}{(n+4)(n+5)}

    So we want An+5A+Bn+4B=-1

    n(A+B)+(5A+4B)=-1+0n

    -> A+B=0 \longleftrightarrow A=-B

    And 5A+4B=-1 \implies 5(-B)+4B=-1 \implies \boxed{B=1} \implies \boxed{A=-1}



    So \frac{-1}{(n+4)(n+5)}=\frac{-1}{n+4}+\frac{1}{n+5}


    This may be better doing this way for you..
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  4. #19
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    last question

    Moo
    on that new way you showed me how would i get the sum?

    i stick values for n of 0, 1, 2, 3... and see if there is number not reapeating?
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  5. #20
    Moo
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    Quote Originally Posted by taurus View Post
    last question

    Moo
    on that new way you showed me how would i get the sum?

    i stick values for n of 0, 1, 2, 3... and see if there is number not reapeating?
    The number which does not repeat is always the first one in a telescoping sum... I gave you a link to wikipedia and TheEmptySet explained it well.

    And n begins at 1
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  6. #21
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    but for the new method seems more understandable
    i put n = 1 and got -1/30 which aint the sum:?
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  7. #22
    Moo
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    It's n=1 in \frac{-1}{n+4}
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  8. #23
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    Question

    but you said
    (-1/n+4) + (1/n+5)
    why dont you add?
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  9. #24
    Moo
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    Because 1/(n+5) annulates with the following term, and so on...

    Just write the few first terms of the series on a paper, and you will see that..
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