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So I have this sum of exponentials and I would like to find an expression for it.

$$\sum^n_{i=1} e^{\mu(i-1)} $$

Note that $i$ is not an imaginary indicator. I am aware there is a formula for summing a purely exponential sum, but I am not clear as to what happens to the $\mu$.

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Define $a=e^\mu$ when $\mu\ne 0$. Then you have $$\sum^n_{i=1} e^{\mu(i-1)} =\sum^n_{i=1} a^{i-1}=1+a+\cdots+a^{n-1}={a^n-1\over a-1}={e^{\mu n}-1\over e^\mu -1} $$For $\mu =0 $ we obtain$$\sum^n_{i=1} e^{\mu(i-1)}=n$$

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The sum $S$ can be rewritten as $$S= \sum_{i=0}^{n-1} e^{\mu i}=\sum_{i=0}^{n-1} (e^{\mu})^i=\frac{1-e^{\mu n}}{1-e^{\mu}}$$ since the geometric series $$\sum_{i=0}^{n-1} x^{i}=\frac{1-x^n}{1-x}$$

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One may recall that $$ \sum_{i=1}^nx^{i-1}=\frac{1-x^n}{1-x},\qquad x\neq1. $$ What if you put $x=e^\mu$?

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Hint:

This the sum of the $n$ first terms of the geometric series with ratio $\mathrm e^\mu$, since $\;\mathrm e^{\mu(i-1)}=(\mathrm e^{\mu})^{i-1}$.

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