How to find sum of factors of $2^{2012}$?

Question

This question really is confusing me and I was wondering if there was a simple way this could be achieved. I've come up with this so far after skimming through a few articles on the net. I assumed this may be correct as the number is already expressed in terms of its prime factors.

$$\sum_{n=0}^{2012} (2^n)$$

Answer 1

Since $2$ is the only prime factor, the sum of the factors of $2^{2012}$ is indeed: $$\sum\limits_{k=0}^{2012} 2^k = 2^0+2^1+2^2+\cdots+2^{2012}\quad\quad\color{red}{\checkmark}$$

Next notice that this is a geometric progression, thus use: $$\sum\limits_{k=0}^{n} ar^k = \dfrac{a(r^{n+1}-1)}{(r-1)}$$

Answer 2

Generally, the sum of the factors of $n = \prod_ip_i^{a_i} $ equals $ \prod_i(1+p_i+p_i^2+ \cdots + p_i^{a_i})$.

Answer 3

Try doing this manually for $2^1$, $2^2$, $2^3$, and $2^4$, and see if you can see the pattern.