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## Topic review (newest first)

MathsIsFun
2005-09-18 08:17:17

Hmmm ... without the power of 1.1 it is fairly easy to calculate the "cumulative sum" - it is the average times the count.

For example, the cumulative sum of 1,2,3,4,5 is the average value (3) times the count (5) = 3x5 = 15, and the average of a series of numbers can be worked out by just averaging the first and last (ie (1+5)/2 = 3)

But the " ^1.1 " makes it a geometric series, not linear ...

So I think we are looking at "Power Sums", a slightly tricky area of mathematics that I don't have enough knowledge of.

Sten
2005-09-18 02:33:08

I've always been terrible at working backwards from a formula.

units = 300 * (growingNumber^1.1)

The above formula determines how many units a certain item requires to complete. Each additional item increases in unit consumption to complete. My task is to determine a formula that will return the total number of items that can be built, given a certain number of units.

ie.
The above formula produces:
Item#  Units
1     300
2     643
3     1,005
4     1,378
5     1,762
6     2,153
7     2,551
8     2,955
9     3,363
10     3,777

If a person has 5008 units, he can then produce 5 total Items (300+643+1005+1378+1762 = 5008).

How can I create a formula to determine that 5 total items can be produced with 5008 units, without actually calculating up every unit cost per item?

Any help would be greatly appreciated.

Regards,
Sten