# Simple geometric series question

1. Jun 16, 2014

### FeynmanFtw

Take the case for the mean:

$$\bar{x} = \frac{1}{N} \Big( \sum_{i=1}^Ni \Big)$$

If we simply use the formula for the sum of a geometric series, we get

$\bar{x} = \frac{N}{2} (2a + (N - 1)d)$

where a and d both equal 1, so we simply get the result

$\bar{x} = \frac{1}{2} (N + 1)$

What I've been trying to do is to get the same result by being more rigorous, in that I've attempted to expand the series, i.e.

$$\bar{x} = \frac{1}{N} (1 + 2 + 3 + ... + N)$$

and replace the terms with the appropriate expressions using N, for example 1 and 2 would be N-(N-1) and N-(N-2) respectively, and so forth. Unfortunately I just keep going round in circles and never achieve the correct result.

Am I wasting my time? Or have I simply not seen the next step?

2. Jun 16, 2014

### micromass

The idea is to not replace all the terms, but only have the terms. For example $2$ gets replaced by $N - (N - 2)$, but we don't want to replace the later term $N-2$. The terms will cancel out then.

3. Jun 16, 2014

### FeynmanFtw

I think I understand, but just to be sure, could you clarify further please?

4. Jun 16, 2014

### micromass

Clarify what? What is unclear about my explanation? Just take your sum $1+2+3+...+N$ and replace the first half of the terms in the way you indicated and leave the other half.

5. Jun 16, 2014

### skiller

I'm still eagerly awaiting the question about a geometric series! :tongue2:

6. Jun 16, 2014

### FeynmanFtw

Well I've already attempted what you've suggested, so it seems, though I cannot reach the final answer, as I've stated already.

I've tried going through the puzzle again and I sometimes obtain N/2 + 1, or N/2 as my final answer instead of (N+1)/2. I'm really confused as to where the +1 comes from. Somehow the final answer evades me.

7. Jun 16, 2014

### micromass

The idea is to do this:

$$1+2+3+4+5 = (5-4) + (5-3) + 3 + 4 + 5 = (5-4) + 4 + (5-3) + 3 + 5 = 5 + 5 + 5= 15$$

8. Jun 17, 2014

### lendav_rott

This looks like an arithmetic sequence to me.

9. Jun 17, 2014

### FeynmanFtw

Well, as is usually the case, I left the problem yesterday evening and decided to sleep on it. This morning I woke up and solved it within a minute. Don't you just love it when you hit a wall and just jump over it the next time you look at a problem from a different angle?

Thanks to all who helped.