$$ L=\frac{\pi t}{8}(\sum_{i=1}^{n}2^{2i}+2\sum_{i=1}^{n}2^{i}) $$ Using the sum formulas for geometric series $$ L=\frac{\pi t}{8}\left(\frac{2^2}{2+1}\frac{2^{2n}-1}{2-1}+4\frac{2^n-1}{2-1}\right)= \frac{\pi t}{8}\left(\frac{4}{3}(2^{2n}-1)+4(2^n-1)\right) $$ Alternate folding is special since folding in one direction doesn't influence the length lost in the other direction (the only impact is in terms of the amount of paper that needs to be folded in the next fold). To derive the formula for alternate folding we start by defining the length of the paper in the two directions as $L_{odd}$ and $L_{even}$. When $n=0$ it's simple: $L_{odd}=L_{even}=W$. After the first fold, as can be seen below: $L_{odd}=\frac{W-\pi t}{2}$ and $L_{even}=W$. After the second fold: $L_{odd}=\frac{W-\pi t}{2}$ and $L_{even}=\frac{W-\pi t -2 \pi t}{2}$.  It's not difficult to conclude that for an arbitrary fold count $n$ if $n$ is even $[L_{even}]_n=\frac{[L_{even}]_{n-1}-\sum^{2^{n-1}}_{j=1}j\pi t}{2} $ and $[L_{odd}]_n=[L_{odd}]_{n-1}$, if $n$ is odd $[L_{odd}]_n=\frac{[L_{odd}]_{n-1}-\sum^{2^{n-1}}_{j=1}j\pi t}{2} $ and $[L_{even}]_n=[L_{even}]_{n-1}$. To understand better how the length is lost as we fold the paper let's take a look at how the paper looks like in the first 3 folds.  For the first fold we see that the lost portion is the one in blue, which corresponds to a semicircle of radius $t$. The length lost is $\pi t$. For the second fold in addition to $\pi t$ we have the portions in red: $\pi t + 2\pi t$ . Finally for the third fold in addition to $\pi t$ and $\pi t + 2\pi t$ we have the portions in green: $\pi t + 2\pi t+ 3\pi t + 4\pi t$ . As we can see every additional fold adds $\sum^{2^{i-1}}_{i=1} j\pi t$ to the total length lost. Britney Gallivan, who was at the time a junior in high school, solved this well-known problem. She was asked by her teacher to fold a sheet of paper 12 times and as an incentive she would get extra credit. She failed multiples times. Later she succeeded after using a thin gold sheet and proved the assumption wrong. Gallivan was able to achieve 12 folds by folding a roll of thin toilet paper that stretched over three-fourths of a mile. It took seven hours in a shopping mall with her parents, but Gallivan was able to bust a myth as well as derive a formula relating the width, thickness of a paper and the number of folds achievable. The urban legend was disproved in 2001. In 2012, students at St. Mark's School in Southborough, Massachusetts visited MIT and beat Gallivan's record with 13 single-direction folds. They didn't actually use Gallivan's single-sheet method, instead choosing to layer the first 64 sheets (equivalent to six folds) on top of each other and then begin the folding. [](https://www.youtube.com/watch?v=vPFnIotfkXo "Paper folding record") As we can see below the function $\frac{1}{6}(2^n+4)(2^n-1)$ grows very rapidly with the number of folds.  Another way to see this is: $$ L=\sum_{i=1}^{n}\sum_{j=1}^{2^{i-1}}j\pi t $$ Now, $\sum_{j=1}^{2^{i-1}}j\pi t$ is a sum of an arithmetic series and so $$ \sum_{j=1}^{2^{i-1}}j\pi t= \frac{\pi t}{2}(2^{2(i-1)}+2^{i-1}) $$ If we don't consider the length lost in radii of earlier folds it means we just have to consider the length lost in the last fold. When we reach the final fold $n$, the side of the square must be at least equal to the length lost in the final fold which is $2^{n-1}\pi t$. Since the total area of the sheet $W^2$ (area = nb of sheets in penultimate step $\times$ area of square in penultimate step) is preserved $$ W^2=2^{n-1}\times (2^{n-1}\pi t)^2 \\ W=\pi t 2^{3/2(n-1)} $$ In the larger perspective, the paper folding experiment connects to exponential growth. This is the essence of exponential growth: very small amounts rapidly become astronomically large through simple doubling. Assuming it was possible to fold paper without restriction, the height of a piece of folded paper would double in thickness each time it was folded. Since one sheet of typical 20-pound paper has a thickness of about 0.1 millimeter, folding 50 times would produce a tower of height $1.13×10^{11}$ meters, and folding one more time would make the stack higher than the distance between the Earth and Sun.