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  1. #1
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    summation

    Evaluate exactly:

    a. <br />
\sum_{k=1}^{100} (1 + (-1)^k)<br />

    b. <br />
\sum_{k=3}^{6} \frac{k^2+2}{k-2}<br />

    For these above do I just plug in the values within the interval into k and then add the total?

    so for b would it be:

    <br />
\frac{3^2 + 2}{3 - 2} + \frac{4^2 + 2}{4 - 2} + \frac{5^2 + 2}{5 - 2} + \frac{6^2 + 2}{6 - 2} = 11 + 9 + 9 + 9.5 = 38.5

    Is this correct or am I way off? haha

    Thank you to anyone who attempts this for me : )
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  2. #2
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    <br />
\sum_{k=1}^{100} (1 + (-1)^k)=0+2+0+2 ... = 100
    As for exercice 2 ... exactly.
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  3. #3
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    Quote Originally Posted by qzno View Post
    Evaluate exactly:

    a. <br />
\sum_{k=1}^{100} (1 + (-1)^k)<br />

    b. <br />
\sum_{k=3}^{6} \frac{k^2+2}{k-2}<br />

    For these above do I just plug in the values within the interval into k and then add the total?

    so for b would it be:

    <br />
\frac{3^2 + 2}{3 - 2} + \frac{4^2 + 2}{4 - 2} + \frac{5^2 + 2}{5 - 2} + \frac{6^2 + 2}{6 - 2} = 11 + 9 + 9 + 9.5 = 38.5

    Is this correct or am I way off? haha

    Thank you to anyone who attempts this for me : )
    For the first one, write out some of the terms and group in pairs (some will cancel).

    The second you have is correct.
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  4. #4
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    Thank you both very much : )
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  5. #5
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    Hello, qzno!

    Evaluate exactly:

    (a)\;\;\sum_{k=1}^{100} (1 + (-1)^k)

    (b)\;\;\sum_{k=3}^{6} \frac{k^2+2}{k-2}


    Do I just plug in the values into k and then add the total? . . . . Yes!

    So for (b) would it be:

     \frac{3^2 + 2}{3 - 2} + \frac{4^2 + 2}{4 - 2} + \frac{5^2 + 2}{5 - 2} + \frac{6^2 + 2}{6 - 2} \:= \:11 + 9 + 9 + 9.5 \:= \:38.5 . . . . Right!

    \text{For (a), we have: }\; \underbrace{0 + 2 + 0 + 2 + \hdots + 0 + 2}_{\text{100 terms}}

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  6. #6
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    Quote Originally Posted by qzno View Post
    Evaluate Exactly



    <br /> <br />
\sum_{k=1}^{100} (1 + (-1)^k)<br /> <br />



    I think this can be separated to form



    <br /> <br />
\sum_{k=1}^{100} 1 + \sum_{k=1}^{100} (-1)^k<br /> <br />



    But I do not know where to go from here

    Are there any rules or formulas to complete this?



    And I know that

    <br /> <br />
\sum_{k=1}^{100} 1 = 100<br /> <br />
    What is the value of (-1)^k for even values of k? For odd values of k?

    What then is the value of 1 + (-1)^k for even values of k? For odd values of k?

    How many even-value pairs do you have? How many odd?

    What then is the sum?
    Last edited by mr fantastic; January 14th 2009 at 02:53 PM. Reason: Added the quote - trying to sort out this double post (replies at both) mess
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  7. #7
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    Quote Originally Posted by qzno View Post
    Evaluate Exactly



    <br /> <br />
\sum_{k=1}^{100} (1 + (-1)^k)<br /> <br />



    I think this can be separated to form



    <br /> <br />
\sum_{k=1}^{100} 1 + \sum_{k=1}^{100} (-1)^k<br /> <br />



    But I do not know where to go from here

    Are there any rules or formulas to complete this?



    And I know that

    <br /> <br />
\sum_{k=1}^{100} 1 = 100<br /> <br />
    You're getting there. Try expanding the sums so you get an idea of what's going on:

     {\color{red}\sum_{k=1}^{100} 1} + {\color{blue}\sum_{k=1}^{100} (-1)^k} = \underbrace{{\color{red}1 + 1 + \cdots + 1}}_{100 \ \text{times}} \ + \ \left[{\color{blue}(-1) + (-1)^2  + \cdots  + (-1)^{99} + (-1)^{100}}\right]

    Recall that: (-1)^n = \begin{cases} 1 & \text{ if } n \text{ is even} \\ -1 & \text{ if } n \text{ is odd} \end{cases}

    So simplifying the blue:

    \begin{aligned} {\color{red}\sum_{k=1}^{100} 1} + {\color{blue}\sum_{k=1}^{100} (-1)^k} & = {\color{red}100} \ + \ \left[{\color{blue}\underbrace{(-1) + 1}_{= 0} + \underbrace{(-1) + 1}_{= 0} + \cdots + \underbrace{(-1) + 1}_{= 0} }\right]\end{aligned}

    I'm sure you can take it from here.
    Last edited by mr fantastic; January 14th 2009 at 02:54 PM. Reason: Added the quote. Trying to sort out this double post mess (replies at both threads).
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  8. #8
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    Quote Originally Posted by qzno View Post
    Evaluate Exactly

    \sum_{k=1}^{100} (1 + (-1)^k)

    [snip]
    i know that the final answer is 100 overall
    because when u put -1 to an even power, it becomes a +1
    and when u put -1 to an odd power, it becomes a -1

    and you have 50 even and 50 odd numbers between 1 and 100
    but i was wondering if there were any formulas or rules for this : )
    Last edited by mr fantastic; January 14th 2009 at 02:57 PM. Reason: Added the quote. Tryng to sort out this double post mess (replies at both threads, OP follow-up questions etc.)
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  9. #9
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    Excellent o_O
    Thanks For Responding Both Of You : )
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