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) ?

A formula for $$r+r^2+r^4+\cdots+r^{2^k}$$ is $$r+r^2+r^4+\cdots+r^{2^k}$$ You can calculate the sum of the series, if $k$ and $r$ are given, by calculating the sum of the series, that is, by computing the individual terms and adding them together.

Presumably, what is desired is a closed form formula for the sum, in terms of familiar functions such as logarithms and exponentials and sines and cosines and powers and roots, and a more efficient way to calculate it. As indicated in the comments, no such formula is known. I don't know whether it is possible to prove that no such formula exists, but considering how elementary the question is, I am confident that if there were a nice formula for it, Euler would have found it 250 years ago, and we'd all know about it now.

I haven’t been able to obtain a closed form expression for the sum, but maybe you or someone else can do something with what follows.

In Blackburn's paper (reference below) there are some manipulations involving the geometric series

$$1 \; + \; r^{2^n} \; + \; r^{2 \cdot 2^n} \; + \; r^{3 \cdot 2^n} \; + \; \ldots \; + \; r^{(m-1) \cdot 2^n}$$

that might give some useful ideas to someone. However, thus far I haven't found the identities or the manipulations in Blackburn's paper to be of any help.

Charles Blackburn, Analytical theorems relating to geometrical series, Philosophical Magazine (3) 6 #33 (March 1835), 196-201.

As for your series, I tried exploiting the factorization of $r^m – 1$ as the product of $r-1$ and $1 + r + r^2 + \ldots + r^{m-1}:$

First, replace each of the terms $r^m$ with $\left(r^{m} - 1 \right) + 1.$

$$S \;\; = \;\; \left(r – 1 \right) + 1 + \left(r^2 – 1 \right) + 1 + \left(r^4 – 1 \right) + 1 + \left(r^8 – 1 \right) + 1 + \ldots + \left(r^{2^k} – 1 \right) + 1$$

Next, replace the $(k+1)$-many additions of $1$ with a single addition of $k+1.$

$$S \;\; = \;\; (k+1) + \left(r – 1 \right) + \left(r^2 – 1 \right) + \left(r^4 – 1 \right) + \left(r^8 – 1 \right) + \ldots + \left(r^{2^k} – 1 \right)$$

Now use the fact that for each $m$ we have $r^m - 1 \; = \; \left(r-1\right) \left(1 + r + r^2 + \ldots + r^{m-1}\right).$

$$S \;\; = \;\; (k+1) + \left(r – 1 \right)\left[1 + \left(1 + r \right) + \left(1 + r + r^2 + r^3 \right) + \ldots + \left(1 + r + \ldots + r^{2^{k} - 1} \right) \right]$$

At this point, let's focus on the expression in square brackets. This expression is equal to

$$\left(k+1\right) \cdot 1 + kr + \left(k-1\right)r^2 + \left(k-1\right)r^3 + \left(k-2\right)r^4 + \ldots + \left(k-2\right)r^7 + \left(k-3\right)r^8 + \ldots + \left(k-3\right)r^{15} + \ldots + \left(k-n\right)r^{2^n} + \dots + \left(k-n\right)r^{2^{n+1}-1} + \ldots + \left(1\right) r^{2^{k-1}} + \ldots + \left(1\right) r^{2^{k} - 1}$$

I'm now at a loss. We can slightly compress this by factoring out common factors for groups of terms such as $(k-2)r^4 + \ldots + (k-2)r^7$ to get $(k-2)r^4\left(1 + r + r^2 + r^3\right).$ Doing this gives the following for the expression in square brackets.

$$\left(k+1\right) + \left(k\right)r + \left(k-1\right)r^2 \left(1+r\right) + \left(k-2\right)r^4\left(1+r+r^2+r^3\right) + \ldots + \;\; \left(1\right)r^{2^{k-1}} \left(1 + r + \ldots + r^{2^{k-1} -1} \right)$$