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I am putting pricing strategy for a product

This product for 1 user is for $100

for 2 user is for $190

for 3 user is for $271

and so on

there for the equation will be

1 user = 100

2 users = $100 + 100\times0.9$

3 users = $100 + 100\times0.9 + 100\times0.9^2$

4 users = $100 + 100\times0.9 + 100\times0.9^2 + 100\times0.9^3$

..

There for for $N$ users the price will be

$N$ users = $100 (1 + 0.9 + 0.9^2 + 0.9^3 + ... + 0.9^N)$

How to put this in a single formula so I do not need to calculate all $N-1$ users first?

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  • $\begingroup$ Are you familiar with the sum of a geometric series? $\endgroup$
    – Toby Mak
    Commented Nov 29, 2020 at 7:16

2 Answers 2

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The sum of the geometric series $1+r+r^2+\ldots+r^N$ is given by $\frac{1-r^{N+1}}{1-r}$ if $r\not=1$ (and by $N+1$ if $r=1$), Thus in your case you may calculate $100\frac{1-0.9^{N+1}}{1-0.9 }=1000({1-0.9^{N+1}}) $.

Comment: As mentioned in a comment below $\frac1{0.1}=10$ and not $100$.

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  • $\begingroup$ 100 / 0.1 = 1000 $\endgroup$
    – asmgx
    Commented Nov 29, 2020 at 7:52
  • $\begingroup$ Of course! My fault, $\endgroup$ Commented Nov 29, 2020 at 8:23
  • $\begingroup$ I believe your $N$ is off by $1$ from the question. (Try $N=1$.) $\endgroup$
    – robjohn
    Commented Nov 29, 2020 at 10:53
  • $\begingroup$ You are right, but the formula in the question was given by $1+0.9+\ldots +0.9^N$, $\endgroup$ Commented Nov 29, 2020 at 15:16
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You can solve it with the geometric formula. Therefore, you get $$ \sum_{k=1}^{n} 100 \cdot 0.9^{k-1} = 100\cdot \frac{1-0.9^{k}}{1-0.9} $$

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  • $\begingroup$ When K = 1 the sigma side = 900 while the right side is 100 !! i think it should be k+1 $\endgroup$
    – asmgx
    Commented Nov 29, 2020 at 7:52
  • $\begingroup$ and still does not cover the case when k = 0 $\endgroup$
    – asmgx
    Commented Nov 29, 2020 at 7:54
  • $\begingroup$ @asmgx You're right, I made an error, though the error was a different one. It should have been $k-1$ on the right-hand side. And I think the case k=0 is not necessary in the example above, because it begins at 1. $\endgroup$
    – KaiL1ng
    Commented Nov 29, 2020 at 9:49

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