# Sum with multiple components

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1. Jun 28, 2016

### 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).

2. Jun 28, 2016

### ehild

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.

3. Jun 28, 2016

### NihalRi

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.

4. Jun 28, 2016

### 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.

5. Jun 28, 2016

### SammyS

Staff Emeritus
@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?

6. Jul 16, 2016

### micromass

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

7. Jul 16, 2016

### Staff: Mentor

I don't see what the difficulty is here. All that's needed here is to add four numbers.

8. Jul 17, 2016

### Rectifier

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