# form of the addition with exponent $m^0+m^1+…+m^{n-1}$

I have a series of addition $m^0+m^1+m^2+…+m^{n-1}$ for natural numbers $m$ and $n$. is there a simplified formula for this? How its related to $m^n$?

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What you have here is a geometric progression –  Old John Jan 10 '14 at 21:37
@OldJohn Thanks I was missing this keyword. The form is $\frac{1-m^n}{1-m}$ –  seteropere Jan 10 '14 at 21:59
You are welcome - glad to help. –  Old John Jan 10 '14 at 22:02

It is actually a geometric series that you posted here. so for a Geometric series(GP series) we have a closed formula. so now let us have a look: if we have a series such as... a + ar + ar^2 + ar^3 + ... + ar^n (note that all the powers are only for 'r' not for 'a') if anyone ask for the sum for this series then we can say that the sum is S= a(((r^n)-1)/(r-1)).note that here 'a' is the first term and the value 'r' is the ration of any two consecutive term of the series.hope this discussion will help you.best of luck.

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Hint: Multiply your sum with $m-1$, and see what you get. :-)

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I got $m^n-1$ :) –  seteropere Jan 13 '14 at 18:08

Hi friend the formula I have given to here as S= a((r^n)-1)/(r-1) it is correct but it holds only when the value of 'r' is greater than 1 but if the value of 'r' is less than 1 or the value of 'r' is a fraction then the formula will be like this(all same but the denominator part will be (1-r)) S= a(((r^n)-1)/(1-r)). sorry friend for my last answer.I should give this formula as well.

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Welcome to math.SE: you can edit your existed answer instead of creating a new one :) –  seteropere Jan 10 '14 at 22:00