show that $\sum^\infty_{i=1}p^2(1-p)^{2(i-1)}=\frac{p^2}{1-(1-p)^2}$

Question

In my homework assignment it says: $\sum^\infty_{i=1}p^2(1-p)^{2(i-1)}=\frac{p^2}{1-(1-p)^2}=\frac{p}{2-p}$

I don't see how $\sum^\infty_{i=1}p^2(1-p)^{2(i-1)}=\frac{p^2}{1-(1-p)^2}$

I tried showing it, knowing that:

$\sum^\infty_{i=1}p^2(1-p)^{2(i-1)} = \frac{p^2}{(1-p)^2}\sum^\infty_{i=1}(1-p)^{2i}=$

$\frac{p^2}{(1-p)^2}\sum^\infty_{i=0}(1-p)^{2i}-(1-p) = \frac{p^2}{(1-p)^2}\sum^\infty_{i=0}((1-p)^{2})^{i}-(1-p) = \frac{p^2}{(1-p)^2} \cdot ( \frac{1}{(1-p)^2} -1 )$

it does not seem to be $\frac{p^2}{1-(1-p)^2}$

is there a mistake I've made along the way? Can someone help?

Answer 1

Let's simplify things a bit first. Let $j=i-1$ giving us

$$\sum_{j=0}^{\infty} p^2 (1-p)^{2j}. $$

Notice there's no dependence on $j$ for the $p^2$ term which enables us to drag it out of the summation

$$p^2\sum_{j=0}^{\infty} (1-p)^{2j}, $$ and to make things clearer, we can make another substitution, say $\tilde{p}=(1-p)^2$, and write

$$p^2\sum_{j=0}^{\infty} \tilde{p}^{j}. $$

Now this is just the geometric series, so we have

$$p^2\sum_{j=0}^{\infty} \tilde{p}^{j} = p^2\left[\frac{1}{1-\tilde{p}}\right] $$ $$= \frac{p^2}{1-(1-p)^2.}$$

Answer 2

The sum of geometric series is

$$\sum_{i=1}^\infty((1-p)^2)^{i-1}=\sum_{i=0}^\infty((1-p)^2)^{i}=\frac{1}{1-(1-p)^2}$$ where we used the fact $\sum\limits_{i=0}^\infty r^i=\frac{1}{1-r}$ for $\vert r\vert<1$.