How to find the correct formula to calculate the sum of a geometric series.

Question

I have a formula for a series which looks like this:

$\sum_{i=2}^{8}(-2)^i$

And I see that this series gives me the following output:

4 - 8 + 16 - 32 + 64 - 128 + 256

Which gives a sum of 172 - assuming I have done the calculations correctly.

However, I also need to find a formula to determine the sum, and I'm wondering if I'm on the right path here:

$\sum_{i=0}^{7}2^i = \frac{2^{7^+1}-1}{2-1} $

This is a formula I found to determine the sum of a geometric series. But I'm a bit thrown off by the task I'm trying to solve above.

I tried approaching the formula as shown below:

$ \frac{(-2)^{6}-1}{-2-1} $

But seing as the formula I have been working on above is a negative integer I'm having some difficulties finding the right approach to calculate the sum with the correct formula.

Am I using the wrong formula or am I simply just using it incorrectly?

Answer 1

We have that for any $r\neq 1$

$$\sum_{i=0}^{n}r^i = \frac{r^{n+1}-1}{r-1}$$

then

$$\sum_{i=0}^{8}(-2)^i = \frac{(-2)^{9}-1}{-2-1}=171$$

and

$$\sum_{i=2}^{8}(-2)^i =\sum_{i=0}^{8}(-2)^i-\sum_{i=0}^{1}(-2)^i=171-(1-2)=172$$

Answer 2

Sometimes you can find close formed solutions for values of sums by noticing patterns. For example, we have $S = \sum_{i=0}^n(-2)^i = \sum_{i=0}^n(-1)^i(2)^i$, but what is $2S$? We can see that $2S = \sum_{i=0}^n(-1)^i(2)^{i+1} = \sum_{i=1}^{n + 1}(-1)^{i-1}(2)^i = \sum_{i=1}^{n + 1}-(-1)^{i}(2)^i = 1 - S + (-1)^n2^{n + 1}$.

So $3S = 1 + (-1)^n2^{n+1} \implies S = \frac{1 + (-1)^n2^{n+1}}{3}$.

You can use this to get the sum in any interval by simply subtracting.