# Solving for the CDF of the Geometric Probability Distribution

So I am trying to find the CDF of the Geometric distribution whose PMF is defined as $$P(X=k) = (1-p)^{k-1}p$$ where X is the number of trials up to and including the first success.

Now attempting to find the general CDF, I first wrote out a few terms of the CDF:

$$P(X=1) = p \\P(X=2) = p(1-p) + p \\ P(X=3) = p(1-p)^2 + p(1-p) + p\\....P(X=k) = p(\sum\limits_{i=1}^{k-1} (1-p)^i)$$

Now I know this last sum has to equal 1, therefore: $$p(\sum\limits_{i=1}^{k-1} (1-p)^i) = 1$$

Now I am aware that the CDF is supposed to be $$F(X=k) = 1-(1-p)^k$$

What I am trying to figure out is how to go from what I have to the final solution. Any hints or ideas? Thanks

The CDF is defined as $$F(k)=P(X\leq k)=\sum_{k'=1}^k P(X=k')=\sum_{k'=1}^k p (1-p)^{k'-1}=1-(1-p)^k\ ,$$ using a finite geometric sum .