# Simplified C.D.F.

1. Mar 16, 2009

### 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?

Last edited: Mar 17, 2009
2. Mar 17, 2009

### 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?

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

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