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Is there a formula that can perform this discounting of a constant cash flow to present value without having to do the whole summation?

$$DCF=\dfrac{CF_1}{(1+r)^1}+\dfrac{CF_2}{(1+r)^2}+\cdots+\dfrac{CF_n}{(1+r)^n}$$

As you can see; currently I have to discount each year's cash flow individually to get its present value. The cash flow each year is the same so is there a formula that I can use rather than punching into my calculator this whole $\frac{1}{(1+i)^1}+\frac{1}{(1+i)^2}+\frac{1}{(1+i)^3}+\frac{1}{(1+i)^n}$

Raynos
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  • But you also have those changing $;F_1,F_2,...,F_n;$ in the numerator... – DonAntonio Dec 07 '13 at 00:06
  • this is a simple geometric progression with $r = \frac{1}{1+i}$ http://en.wikipedia.org/wiki/Geometric_progression – abkds Dec 07 '13 at 00:06
  • I just said that C and F, the numerator will always be a constant..I need a way to summarize that 1 over bottom portion.. – Raynos Dec 07 '13 at 00:11

1 Answers1

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In the equation you indicate that the cash flows are different each year, in which case there is nothing to be done. In the text you indicate that the flows are all the same, in which case you can distribute it out and get a geometric series. $$DCF=CF\left((1+r)^{-1}+(1+r)^{-2}+\dots (1+r)^{-n}\right)\\=CF\left(\frac {(1+r)^{-1}-(1+r)^{-n-1}}{1-(1+r)^{-1}}\right)\\=CF\frac {1-(1+r)^{-n}}{r}$$

Ross Millikan
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