<s>[INST] Solving a partial sum.... Hey can anyone help with this? This is the classic NPV equation:
$$\texttt{NPV = -CapEx}  +  \sum_{i=0}^n \frac{\texttt{Revenue − Costs}}{(1+\texttt{Discount})^i}$$
For my purposes all the elements are know except costs.
I need to isolate costs in this equation. Is this possible?
Thanks,
Mike
 [/INST] Your equation is 
$NPV = -CapEx + \sum_{i=0}^n \frac{R − C}{(1+d)^i} \quad |+CapEX$
$NPV + CapEx = \sum_{i=0}^n \frac{R − C}{(1+d)^i} \quad $
Splitting the fraction
$NPV + CapEx = \sum_{i=0}^n \left( \frac{R }{(1+d)^i} -\frac{C }{(1+d)^i} \right) \quad $
Factoring out $\frac{1}{(1+d)^i}$
$NPV + CapEx = \sum_{i=0}^n \frac{1}{(1+d)^i}\left( R-C \right) \quad $
$NPV + CapEx = \left( R-C \right)\cdot \sum_{i=0}^n \frac{1}{(1+d)^i} \quad | \cdot (-1)$
$-NPV - CapEx = \left( -R+C \right)\cdot \sum_{i=0}^n \frac{1}{(1+d)^i} \quad $
Dividing the equation by $\sum_{i=0}^n \frac{1}{(1+d)^i}$
$\frac{-NPV - CapEx}{\sum_{i=0}^n \frac{1}{(1+d)^i}}=-R+C$
$C=R-\frac{NPV + CapEx}{\sum_{i=0}^n \frac{1}{(1+d)^i}}$
It is $\sum_{i=0}^n \frac{1}{(1+d)^i}=\sum_{i=0}^n \left( \frac{1}{1+d}\right)^ i$
This is a partial sum of a geometric series. 
Therefore $\sum_{i=0}^n \left( \frac{1}{1+d}\right)^ i=\Large{\frac{1-\left( \frac{1}{1+d}\right)^{n+1}}{1-\frac{1}{1+d}}}$
In total it is $C=R-(NPV+CapEx)\cdot \Large{\frac{1- \frac{1}{1+d}}{1-\left(\frac{1}{1+d}\right)^{n+1}}}$
Solve for R-C
$R-C=(NVP+CapEx)\cdot \Large{\frac{\left( 1- \frac{1}{1+d}\right) }{\left(1-\left(\frac{1}{1+d}\right)^{n+1}\right)}}$
$R-C=(1909+1315)\cdot \Large{\frac{\left(1- \frac{1}{1+0.08}\right)}{\left(1-\left(\frac{1}{1+0.08}\right)^{24}\right)}}=\normalsize 283.5268$
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