# How to calculate $k^0+k^1+k^2 + k^3+…+ k^{n-1}$ [duplicate]

How to simplify below expression or convert it to something simpler like $k^{n-1}$?

$$k^0+k^1+k^2 + k^3+...+ k^{n-1}$$

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## marked as duplicate by Did, drhab, Ayman Hourieh, Raff, Adam HughesJul 28 at 15:57

What happens when you multiply your expression by $k$? –  Mark Bennet Jul 28 at 15:26
This is a finite geometric series. You have probably seen it before, maybe as $1+r+r^2+\cdots +r^{n-1}$. –  André Nicolas Jul 28 at 15:28

Let $\displaystyle S=k^0+k^1+k^2+\cdots +k^{n-2}+k^{n-1}\ \ \ \ (1)$

If $k=1, S=n$

Else

$\displaystyle k\cdot S=k^1+k^2+k^3+\cdots +k^{n-1}+k^{n}\ \ \ \ (2)$

Now subtract $(2)$ from $(1)$ to find $\displaystyle (k-1)S=k^n-1$

Reference : Geometric progression

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This is a geometric series. If $k \neq 1$, it's sum can be derived as follows:

$S = k^0+k^1+k^2+\cdots+k^{n-1}$

$kS = k^1+k^2+k^3+\cdots+k^{n}$

$S-kS = k^0 - k^n$

$(1-k)S = 1-k^n$

$S = \dfrac{1-k^n}{1-k}$

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For $k \neq 1$:

$$\sum_{i=0}^{n-1} k^i=\frac{k^n-1}{k-1}$$

For $k=1$:

$$\sum_{i=0}^{n-1} 1=n$$

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@evinda what will be the result if $k=1?$ –  lab bhattacharjee Jul 28 at 15:28
@labbhattacharjee I edited my answer!!! –  evinda Jul 28 at 15:36

Hint: Let $S$ be the given sum, then find $kS$, and subtract the first from the second telescopingly...

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