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...

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

$$\alpha \frac{1-(1-\alpha)^{n+1}}{1-(1-\alpha)}$$


$$\alpha \frac{-(1-\alpha)^{n+1}}{-1+\alpha}$$

Is this right?


$$\frac{-(1-\alpha)^{n+1}}{-1}$$


$$(1-\alpha)^{n+1}$$