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The series $\frac{1}{2^k}$ is a geometric progression. Hence, $\sum_{k=0}^m \frac{1}{2^k} = \frac{1-\left(\frac{1}{2}\right)^m}{1-\left(\frac{1}{2}\right)}$ You can continue from here.


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Try to prove instead (without induction) that if the algorithm returned false, there is no hamilton path. Hint: there must be two parallel nodes. Can we reach one from the other?


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