Closed form to series

Question

Is there a closed form of the series

$$ \sum_{d=0}^D F^d $$

where D is a finite integer, not $\infty$, like in power series?

Answer 1

The method for demonstrating the formula given by Bernard in the comments above is called telescoping. The trick is to observe that the sum can be written as:

$$ S \;\; =\;\; \sum_{d=0}^D F^d \;\; =\;\; 1 + F + F^2 + \ldots + F^D. $$

If I multiply $S$ by an additional copy of $F$ we obtain:

$$ FS \;\; =\;\; F\left (1 + F + F^2 + \ldots + F^D \right ) \;\; =\;\; F + F^2 + \ldots + F^D + F^{D+1}. $$

We can now subtract one equation from the other to obtain the telescoping effect: \begin{eqnarray*} S - FS & = & \left (1 + F + F^2 + \ldots + F^D \right ) - \left( F + F^2 + \ldots + F^D + F^{D+1} \right ) \\ & = & 1 - F^{D+1}. \end{eqnarray*}

Notice that the above subtraction eliminates all but the first and last terms. Acknowledging that $S$ was the sum we obtained, we can isolate it by noticing that $S - FS = S(1-F)$. Dividing both sides we find:

$$ S \;\; =\;\; \sum_{d=0}^D F^d \;\; =\;\; \frac{1 - F^{D+1}}{1-F}. $$