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Sum with multiple components

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  1. Jun 28, 2016 #1

    Rectifier

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    The problem
    I want to calculate the following sum
    $$ \sum^{5}_{k=2} \frac{k(-1)^k}{2^k} $$

    The attempt

    I wrote ## \frac{(-1)^k}{2^k} ## as ##\frac{1}{(-2)^k}##. I was hoping that I could calculate the sum ## \sum^{5}_{k=2} \frac{k(-1)^k}{2^k} ## by multiplying the sums ##\sum^{5}_{k=2} k## and ## \sum^{5}_{k=2} \frac{1}{(-2)^k}##, but I was wrong (wolfram).
     
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  3. Jun 28, 2016 #2

    ehild

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    What is the sense to multiplying the sum with an other sum????
    Just write the terms from k=2 to k=5 and add them.
     
  4. Jun 28, 2016 #3
    This part looks fine to me.
    Expanding thus sumation would give
    (2 (-1)^2)/2^2 +(3 (-1)^3)/2^3 ... (5(-1)^5)/2^5

    You can see here that we can not separate out ∑ k as a multiplication factor. Therfore you cannot calculate the two sums separately then multiply them. I think if you are allowed a calculator it would be easier to just substitute in the numbers into the expansion.
     
  5. Jun 28, 2016 #4

    Rectifier

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    I am not allowed to use a calculator. It looks like we are supposed to add these manually without using any specific formula. Thanks for your help.
     
  6. Jun 28, 2016 #5

    SammyS

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    @Rectifier ,

    Let's put NihalRi's expression into LaTeX and do some very basic simplifying.
    ##\displaystyle \frac{2 (-1)^2}{2^2} +\frac{3 (-1)^3}{2^3}+\frac{4 (-1)^4}{2^4} +\frac{5 (-1)^5}{2^5} \ ##

    ##\displaystyle \frac{2 }{4} -\frac{3 }{8}+\frac{4 }{16} -\frac{5 }{32} \ ##​

    It's not that difficult without a calculator, now is it?
     
  7. Jul 16, 2016 #6

    micromass

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    I agree that just writing out the terms and adding them is the best approach. If you want a general formula however, then take the derivative of ##\sum_{k=1}^n x^k##.
     
  8. Jul 16, 2016 #7

    Mark44

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    I don't see what the difficulty is here. All that's needed here is to add four numbers.
     
  9. Jul 17, 2016 #8

    Rectifier

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    Thank you for commenting but I have marked this question as solved some time ago.
     
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