Simplified C.D.F. Formula: Get the Answer Now!

[needhelp83]

Derive a simplified formula for the c.d.f., F(n) = (P \leq n), using:

\sum_{k=0}^{n}r^k=\frac{1-r^{n+1}}{1-r}


p(n)=\alpha (1-\alpha)^n

\alpha \sum_{k=0}^{n}r^k=\frac{1-(1-\alpha)^{n+1}}{1-(1-\alpha)}

How do I break this down to a simplified version?

 

[needhelp83]

Alright, I may have figured this out hopefully...

<br /> \alpha \sum_{k=0}^{n}r^k=\frac{1-(1-\alpha)^{n+1}}{1-(1-\alpha)} <br />

<br /> \alpha \frac{1-(1-\alpha)^{n+1}}{1-(1-\alpha)} <br />


<br /> \alpha \frac{-(1-\alpha)^{n+1}}{-1+\alpha} <br />

Is this right?


<br /> \frac{-(1-\alpha)^{n+1}}{-1} <br />


<br /> (1-\alpha)^{n+1}<br />