I am having a hard time researching how to handle summations of functions with exponential growth or decay. I know that simple summations can be calculated as follows:
$$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$$
How do you approach problems of exponential decay or growth? Consider the following example:
$$\sum_{n=1}^{50} e^{-0.123(n)}$$
On
\begin{align} S & = \sum_{n=1}^{50} e^{-0.123n} & & = \phantom{a_0 + {}} a_1 + a_2 + \cdots + a_{49} + a_{50} \\[10pt] \frac S {e^{-0.123}} & = \sum_{n=0}^{49} e^{-0.123n} & & = a_0 + a_1 + a_2 + \cdots + a_{49} \\[10pt] \end{align} Subtracting the first row above from the second we get $$ \frac S {e^{-0.123}} - S = a_0 - a_{50}. $$ Therefore $$ S\left( 1 - \frac 1 {e^{-0.123}} \right) = a_0 - a_{50} $$ $$ S = \frac{a_0-a_{50}}{1 - \frac 1 {e^{-0.123}}} = \frac {1 - e^{-0.123(50)}}{1- \frac 1 {e^{-0.123}}} $$
You can recognize your sum as a geometric sum which has the basic formula: $$ \sum_{n=0}^N r^n = \frac{r^{N+1} - 1}{r-1} $$ To apply this to your sum $$ \sum_{n=1}^{50} e^{-0.123(n)} $$ recognize that $e^{-0.123(n)} = (e^{-0.123})^n$ so your $r$ is $e^{-0.123}$.
Also your sum starts at $n=1$ while the formula starts the summation at $n=0$ so you need to adjust for this.