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What is the formula for n*1+n*2+n*3+n*4...?

Josiah Bertoli
Posted 3 years ago

I'm not even sure what section of mathematics this is to be able to research it myself, please point me in the right direction! It might be set theory?

I'm trying to create a formula that can give the monthly cost of a unit of storage in a system with the assumption that the system will become fully utilised at a set rate and each unit added remains in the system. Before I can get to a cost, I need to work out a percentage per unit share of the total cost.

For example, at a rate of 10 additional units per month being added to the storage, having a storage lifetime of 1 year, and having a starting number of units in storage of 0, the utilisation of the storage could be represented as the following:

10*1+10*2+10*3+10*4+10*5+10*6+10*7+10*8+10*9+10*10+10*11+10*12=780

My question is: how do I turn this into a formula that contains the following variables:

  • x= the number of additional units
  • y= the number of times units are added in a period
  • z= the number of periods
  • BONUS VARIABLE: m= average lifetime of each unit in the system (i.e. how long a unit is stored once added)

This isn't as simple as multiplying all three together and I'm struggling to understand the relationship between each one!

Any help much appreciated!
POSTED BY: Josiah Bertoli
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In[5]:= Sum[n i, {i, 1, m}]

Out[5]= 1/2 m (1 + m) n

Robert
POSTED BY: Robert Nowak
Josiah Bertoli
Josiah Bertoli
Posted 3 years ago

Thanks Rob, very similar to Hans' workings.
POSTED BY: Josiah Bertoli

If your sum is finite, (and if it it this what you mean )

sum = a*1 + a*2 +a*3 +.....+ a*k
sum = a* ( 1 + 2 + ..... + k  )
sum = a* k *(k + 1 ) / 2
POSTED BY: Hans Dolhaine
Josiah Bertoli
Josiah Bertoli
Posted 3 years ago

Perfect Hans, that's exactly what I needed!
POSTED BY: Josiah Bertoli
鹏飞 杜
鹏飞 杜
Posted 3 years ago

I think it involves the concrete mathematics proposed by Knuth and Graham, where you want to translate this summation into closed form, but some summations are not in closed form.Fortunately, your summation has a closed form：

$$ \sum_{k=0}^{n} (a+bk) = a(n+1) + \frac{b(n+1)n}{2} $$

For exmaple，when a=0, b=10, k from 1 to 12, we have

$$ \sum_{k=1}^{12} 10k = \frac{10(12+1)12}{2} = 780 $$
POSTED BY: 鹏飞 杜
Updating Name
Updating Name
Posted 3 years ago

Thanks mate, the extra info about Knuth and Graham is super helpful!
POSTED BY: Updating Name
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