Easiest case. One move or 2^1 - 1.
2^1 - 1 First use the one disk solution to get the top disk
to the center,
1 move the bottom disk to final location,
2^1 - 1 use the one disk solution to get the top disk
to the right.
2^2 - 1 Is the sum.
2^2 - 1 First use the two disk solution to get the top disks
to the center,
1 move the bottom disk to final location,
2^2 - 1 use the two disk solution to get the top disks
to the right.
2^3 - 1 Is the sum.
Non-Math, non-CS majors can simply take the following at face value. Majors will be able to prove the following by induction someday.
2^(N-1) - 1 First use the N-1 disk solution to get
the top disks to the center,
1 move the bottom disk to final location,
2^(N-1) - 1 use the N-1 disk solution to get the top disks
to the right.
2^N - 1 Is the sum.
A table of approximate number of moves:
Disks Moves (est) Moves (act) 10 1,000 1,023 20 1,000,000 1,048,575 30 1,000,000,000 1,073,741,823 40 1,000,000,000,000 50 1,000,000,000,000,000 60 1,000,000,000,000,000,000
For fun, assume that your computer averaged one microsecond (0.000001) for each move. It would take 34,865 years to move all the disks using this solution.