# Thread: How to express / simplify f(x) = x + (x-1) + (x-2) ... (x - x)

1. ## How to express / simplify f(x) = x + (x-1) + (x-2) ... (x - x)

I've forgotten my math...

Is there an easier way to express the function in my title? Here's the context:

In Simcity you can buy a coal plant for $10000. Every year you own it it costs$1 per month * number of years owned * 12 months to maintain. If I've owned the plant for 20 years, it costed me $240 that year,$228 the previous year, $216 the year before that, etc. If I buy a new plant, the maintenance costs go back down to$12 for the year again. So what's the most affordable way to keep using coal plants?

I was thinking, let x = number of years between buying a new plant, and then I could come up with a function that tells me what the cost would be for every x years. It would be something like 10,000 + 12x + 12(x-1) + 12(x-2) and so on, but that's not so easy to evaluate for larger numbers.

So is there a way to simplify that expression? (Or should I just go wind power)?

2. You can use sigma notation to simplify the title function.

3. I guess what I meant was that it would be tedious to evaluate by hand (or brain) for large numbers. I was hoping that was some easier way to express it using only a few terms.

If there isn't, I'll accept that.

4. Originally Posted by supaju
I've forgotten my math...

Is there an easier way to express the function in my title? Here's the context:

In Simcity you can buy a coal plant for $10000. Every year you own it it costs$1 per month * number of years owned * 12 months to maintain. If I've owned the plant for 20 years, it costed me $240 that year,$228 the previous year, $216 the year before that, etc. If I buy a new plant, the maintenance costs go back down to$12 for the year again. So what's the most affordable way to keep using coal plants?

I was thinking, let x = number of years between buying a new plant, and then I could come up with a function that tells me what the cost would be for every x years. It would be something like 10,000 + 12x + 12(x-1) + 12(x-2) and so on, but that's not so easy to evaluate for larger numbers.

So is there a way to simplify that expression? (Or should I just go wind power)?

1. As far as I understand your question you are going to calculate a sum with n+1 summands where n summands are terms in brackets. Right?

2. If so I'm going to modify your notation a little bit:

$s(x) = x+(x-1)+(x-2)+(x-3)+(x-5)+(x-6)+...+(x-n)$

$s(x)=\underbrace{x+x+x+x+...+x}_{\text{(n+1) summands}} - (\underbrace{1+2+3+4+5+6+...+n}_{\text{(n) summands}})$

3. The sum of the first n natural numbers is calculated by:

$1+2+3+...+n=\sum_{i=1}^n(i)=\dfrac n2(n+1)$

$s(x)=(n+1) x - \dfrac n2(n+1)$

5. ## Yes!

Yes, that's the one! Sum of natural numbers...

Thank you earboth!