Sum with multiple components

  • #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).
 

Answers and Replies

  • #2
ehild
Homework Helper
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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).
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.
 
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  • #3
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I wrote (−1)k2k(−1)k2k \frac{(-1)^k}{2^k} as 1(−2)k1(−2)k\frac{1}{(-2)^k}
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.
 
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  • #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.
 
  • #5
SammyS
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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.
@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?
 
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  • #6
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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##.
 
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  • #7
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The problem
I want to calculate the following sum
$$ \sum^{5}_{k=2} \frac{k(-1)^k}{2^k} $$
I don't see what the difficulty is here. All that's needed here is to add four numbers.
Rectifier said:
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).
 
  • #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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