Computing $2^{2^1}+2^{2^2}+2^{2^3}+\cdots+2^{2^n}$ How can I compute the following sum?
$$2^{2^1}+2^{2^2}+2^{2^3}+\cdots+2^{2^n}$$
My attempt was to apply the known formula for the sum of an geometric progression, but it seems that the ratio is variable. So there is a formula for this type of sum?
 A: There is no known closed form for this series.
Anyway, as the terms grow extremely fast, even for moderate $n$,
$$\frac{t_{n-1}}{t_n}=2^{2^{n-1}-2^n}=2^{-2^{n-1}}$$ is a tiny ratio and
$$2^{2^1}+2^{2^2}+2^{2^3}+\ldots+2^{2^n} \approx 2^{2^n}.$$
E.g., for $n=6$, the ratio is $2.3\cdot10^{-10}$. In fact, when keeping only the final term, the value is exact on the first half of the digits.

$$2,4,16,256,65536,4294967296,18446744073709551616,340282366920938463463374607431768211456,\cdots$$
compared to
$$2,6,22,278,65814,4295033110,18446744078004584726,340282366920938463481821351509772796182,\cdots$$
A: Allow me to suggest a paper that discusses what constitutes an answer to these kinds of problems (he focuses on counting problems, but the general idea can be generalized): Herbert S. Wilf, What is an Answer?, The American Mathematical Monthly, Vol. 89, No. 5 (May, 1982), pp. 289-292. The general idea applied to this particular problem would be whether there is a more efficient way to find the sum that just adding term by term.  Working in binary you could express the sum by a string of 0's and 1's in which all digits are zero except for a 1 in positions $2^1$, $2^2$, $2^4$, etc., but that already requires a number of operations at least as long as the number of terms in the sum, so it looks like we are out of luck.
A: Please do not take it as an answer, its just an observation. I am not sure of the consequences but I have solved lots of problem with this method.
Denote, $\displaystyle P_n=\sum_{k=1}^n2^{2^k}$
Note that, $$\sum_{k=0}^{2^n}2^k=\frac{2^{2^n}-1}{2^n-1}$$
$$P_n+Q_n=\frac{2^{2^n}-1}{2^n-1}$$ Where $\displaystyle Q_n=\sum_{k=0}^{2^n}2^k$ where $k\ne2^p$ for any $p>1$
I have no idea how to go for $Q_n$.
A: Let $$\alpha_n = 2^{2^{n}},$$
$$S_n = \sum_{k=1}^{n} \alpha_n.$$
Then
$$ \frac{\alpha_{n+1}}{\alpha_{n}} = \alpha_{n},$$
$$ \alpha_{n+1} = \alpha_{n}^2.$$
$$S_n^2 = (\sum_{k=1}^{n} \alpha_k)^2 =\sum_{k=1}^{n} \alpha_k^2 + 2 \sum_{i <j}^{n} \alpha_{i} \alpha_{j} =\sum_{k=1}^{n} \alpha_{k+1}+2 \sum_{i <j}^{n} \alpha_{i} \alpha_{j}.$$
Fixing the summation
$$S_n^2 = S_n +\alpha_{n+1}-\alpha_{1} +2 \sum_{i <j}^{n} \alpha_{i} \alpha_{j},$$
$$ S_n^2 -S_n +(\alpha_1-\alpha_{n+1}-2 \sum_{i <j}^{n} \alpha_{i} \alpha_{j})=0,  $$
$$ S_n = \frac{1+ \sqrt{8  \sum_{i <j}^{n} \alpha_{i} \alpha_{j}+4 \alpha_{n+1}+1-4\alpha_1}}{2}.$$
The problem is now to calculate 
$$\sum_{i <j}^{n} \alpha_{i} \alpha_{j}.$$
