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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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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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  • 2
    $\begingroup$ @evinda what will be the result if $k=1?$ $\endgroup$ – lab bhattacharjee Jul 28 '14 at 15:28
  • $\begingroup$ @labbhattacharjee I edited my answer!!! $\endgroup$ – evinda Jul 28 '14 at 15:36
1
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Hint: Let $S$ be the given sum, then find $kS$, and subtract the first from the second telescopingly...

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