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Brainstorming answer to a brainstorming question… I'm not happy with this answer myself, and hope that me posting this answer doesn't prevent someone likely to give a better answer from reading your question in the first place. You could start by identifying two points with maximal distance from one another, which you can obtain in linear time from the ...

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We take advantage of your list. Note that there are $1+2+3+\cdots +9$ numbers until the end of the $9$'s. The sum $1+2+3+\cdots+9$ is an arithmetic sequence, with sum $\frac{(9)(10)}{2}$. More generally, the number of numbers until the end of the $n$'s is $\frac{n(n+1)}{2}$. The approximate size of the $N$-th number in our list is the $n$ such that ...

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The answer is to gain experience by studying patterns in other problems. You learn to recognize patterns through practice. It is okay if you don't recognize how to find the pattern in this example at first, but now that you have been shown the pattern, study it carefully and try to see if you can understand why the pattern emerged and how, in hindsight, ...

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Well the only remaining thing to notice is that the nominator is one less than twice the denominator. Then you can write $$\frac{2\cdot2^n-1}{2^n} = 2 - \frac{1}{2^n}$$ and prove it by induction.

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