I'm trying to find a formula that will allow me to calculate the sum total of a progression (not sure if that's the word) in a spreadsheet.

$$1 + 0.79 + 0.79\cdot 0.79 + 0.79\cdot 0.79\cdot 0.79 +\cdots{}$$

I can simplify the parts between the plus signs and graph them with $0.79^x$

But how do I calculate the sum of them without having to first calculate each part. It looks like it should be easy, perhaps similar to $1+2+3+4+5+6 = \frac{6}{2}\cdot 7 = 21$ but this seem to only work with whole numbers.

P.S. The reason I want to do this is to set prices for jobs. $100%$ charge for one hour, $179%$ for two, 241.41% for three, etc. I want to be able to enter in that I worked $3.25$ hrs and get the total price to charge.

Thanks for your help :-)

  • $\begingroup$ Is this an infinite sum? $\endgroup$ – user258700 Dec 5 '15 at 15:17
  • $\begingroup$ $1+x+x^2+x^3+x^4+\cdots=\frac1{1-x}$. $\endgroup$ – Kay K. Dec 5 '15 at 15:25

For every real number $p$, with $0 < p < 1$, the following holds

$$ \sum_{i=0}^{\infty} p^{i} = \frac{1}{1 - p} $$

Hence, if you want to calculate the sum of the first $n$ terms, one has

$$ \sum_{i=0}^{n-1} p^{i} = \frac{1}{1-p} - \sum_{i=n}^{\infty} p^{i} = \frac{1}{1-p} - \frac{p^{n}}{1 - p} = \frac{1 - p^{n}}{1 - p} . $$

For more information, see the wikipedia page on geometric progression

  • $\begingroup$ Thank you! The wiki page explained it to me. I was wanting the formula for a 'geometric series'. (a(1-r^m))/(1-r) $\endgroup$ – Jesse the Wind Wanderer Dec 6 '15 at 10:43

This is called a geometric series. In notation, it can be written as $$\sum_{i = 0}^\infty (.79)^i = \frac{1}{1-.79}.$$ Maybe you recognize that. Have you tried looking that up?


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