Efficient Calculation of a Complex Sum with Multiple Components

[Rectifier]

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]

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.

 

[NihalRi]

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]

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]

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?

 

[micromass]

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 $\sum_{k=1}^n x^k$.

 

[Mark44]

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.

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).

 

[Rectifier]

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