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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Sign up to join this communityHow to simplify below expression or convert it to something simpler like $k^{n-1}$?
$$ k^0+k^1+k^2 + k^3+...+ k^{n-1} $$
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
This is a geometric series. If $k \neq 1$, it's sum can be derived as follows:
$S = k^0+k^1+k^2+\cdots+k^{n-1}$
$kS = k^1+k^2+k^3+\cdots+k^{n}$
$S-kS = k^0 - k^n$
$(1-k)S = 1-k^n$
$S = \dfrac{1-k^n}{1-k}$
For $k \neq 1$:
$$\sum_{i=0}^{n-1} k^i=\frac{k^n-1}{k-1}$$
For $k=1$:
$$\sum_{i=0}^{n-1} 1=n$$
Hint: Let $S$ be the given sum, then find $kS$, and subtract the first from the second telescopingly...