Sum of the form $r+r^2+r^4+\dots+r^{2^k} = \sum_{i=1}^k r^{2^i}$ I am wondering if there exists any formula for the following power series :
$$S = r + r^2 + r^4 + r^8 + r^{16} + r^{32} + ...... + r^{2^k}$$
Is there any way to calculate the sum of above series (if $k$ is given) ?

I have not seen a closed form formula for the sum you talk about, and sense everyone else has already remarked on this fact, I thought I might mention that I have seen several other interesting results involving double exponentials and powers of two if your interested:
$$\sum_{k=0}^{n-1}\frac{2^k}{x^{2^k}+1}=\frac{1}{x-1}-\frac{2^n}{x^{2^n}-1}$$
$$\prod_{k=0}^{n-1}(x^{2^k}+1)=\frac{x^{2^n}-1}{x-1}$$
$$\sum_{k=1}^{2^n}\binom{2^n}{k}=2^{2^n}-1$$
$$\prod_{n=0}^\infty(1+x^{2^n})=\frac{1}{1-x}$$
$$\sum_{n=0}^\infty \frac{2^n}{x^{2^n}+1}=\frac{1}{x-1}$$
$$\sum_{n=0}^\infty\frac{4^nx^{2^n}}{(x^{2^n}+1)^2}=\frac{x}{(x-1)^2}$$
$$\text{Also for |x| < 1}$$
$$\sum_{n=0}^\infty \frac{(-1)^n2^nx^{2^n}}{x^{2^{n+1}}-x^{2^n}+1}=\frac{-2x^2}{x^4+x^2+1} $$
$$\text{Let $f(n)$ count how many factors of 2 n has, then we have} $$
$$\frac{1}{x-1}-\sum_{n=1}^\infty \frac{f(n)}{x^n}=\sum_{n=0}^\infty \frac{1}{x^{2^n}+1}$$
$$\sum_{n=1}^\infty \frac{f(n)}{n^2}=\frac{\pi^2}{18}$$
$$\sum_{k=1}^{2^n} f(k)=2^n-1$$
$$\text{It was also proved by Erdős that both the constants bellow are irrational}$$
$$\sum_{n=1}^\infty \frac{1}{2^n-1}=C_1$$
$$\sum_{n=1}^\infty \frac{1}{2^{2^n}+1}=C_2$$
$$\text{The constant $C_1$ is known as the Erdős-Borwein Constant}$$
$$\text{The constant $C_2$ is less known and is talked about here:}$$
$$\text{http://journal.ms.u-tokyo.ac.jp/pdf/jms080206.pdf}$$
