I was trying to create closed formulas for certain iterated sequences, and came across this sum: $$A := \sum_{k=0}^n q^{2^k} = q+q^2+q^4+\dots+q^{2^n}, \quad q\in\Bbb{R},n\in\Bbb{N}^+$$ I'm trying to find a closed formula for this expression, or a formula where you can obtain the value of the sum by using much fewer arithmetic operations.
I know that the sum
$$S := \sum_{k=0}^n q^k = 1+q+q^2+\dots+q^n$$
has a closed form, since
$$\begin{align} S &= 1+q+q^2+\dots+q^n \\ qS &= q+q^2+q^3+\dots+q^{n+1} \\ qS-S &= q^{n+1}-1 \\ (q-1)S &= q^{n+1}-1 \\ S &= \begin{cases} \frac{q^n-1}{q-1} \quad \text{ if } q \ne 1\\[2ex] n+1 \quad \text{ if } q = 1\end{cases} \end{align}$$
But when trying to use a similar method to determine the value of $A$, we just get a tautology:
$$\begin{align}A = q+q^2+q^4+\dots+q^{2^n}\end{align}$$
Multiplying with $q^c$ won't result in any sort of simplification, so it makes the most sense, to square both sides:
$$\begin{align} A^2 &= \begin{matrix}q^2+q^4+q^8+\dots+q^{2^{n+1}} + q(q^2+q^3+\dots+q^{2^n}) + \\ + q^2(q+q^3+\dots+q^{2^n})+\dots+q^{2^n}(1+q+\dots+q^{2^{n-1}})\end{matrix} \\ A^2 &= A-q+q^{2^{n+1}}+q(A-q)+q^2(A-q^2)+\dots+q^{2^n}(A-q^{2^n}) \\ A^2 &= A-q+q^{2^{n+1}}+qA-q^2+q^2A-q^4+\dots+q^{2^n}A-q^{2^{n+1}} \\ A^2 &= A-q+q^{2^{n+1}}+A(q+q^2+\dots+q^{2^n})-(q^2+q^4+\dots+q^{2^{n+1}}) \\ A^2 &= A-q+q^{2^{n+1}}+A^2-(A-q+q^{2^{n+1}}) \\ A^2 &= A^2 \dots \end{align}$$
I also tried writing up $A$ in the $q$-number system, such that:
$$A = \left(\stackrel{q^{2^n}}{1}\ \underbrace{00\dots 0}_{n-1}\ \stackrel{q^{2^{n-1}}}{1}\ \underbrace{00\dots 0}_{n-2}\ \stackrel{q^{2^{n-2}}}{1} \ \dots \ \stackrel{q^{2^2}}{1}\underbrace{\stackrel{q^3}{0}}_{1}\stackrel{q^{2^1}}{1}\stackrel{q^{2^0}}{1}\stackrel{q^0}{0}\right)_{q}$$
And tried to use formulas to fill up the $0$'s, and relate $A$ to $(11\dots 1)_{q}$, but I just got even more complicated equations that would be more difficult to write up in a closed formula than $A$.
I'm doubtful of there being such a formula, but maybe we can actually obtain one using a different approach.
WolframAlpha also couldn't simplify it, however, that still doesn't mean that such a closed formula doesn't exist.
I'm aware of the trick that
$$A = q+q^2+q^4+\dots+q^{2^n} = q(1+q(1+q^2(1+q^4(1+\dots(1+q^{2^{n-1}})\dots),$$
which involves fewer arithmetic operations to calculate, however I'm only satisfied with a closed formula if it doesn't include dots and/or sum/product/etc. notations. If you could help me find one or show that it doesn't exist using our basic arithmetic notations, that'd be highly appreciated! Thank you in advance.
