- #1
Rectifier
Gold Member
- 313
- 4
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).
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).