Sum with multiple components

Gold Member
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).

ehild
Homework Helper
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.

Rectifier
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.

Rectifier
Gold Member
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.

SammyS
Staff Emeritus
Homework Helper
Gold Member
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?

Rectifier
micromass
Staff Emeritus
Homework Helper
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##.

Irene Kaminkowa and Rectifier
Mark44
Mentor
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).

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