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A small tip: if you want to do this trick, with an exponential F, you probably want a linear search rather than a binary search.

If your k << N then F(N/2) is going to eat up all of the running time of ΣF(i).

Is it faster to start by doubling?

It depends on F, but usually, if F is like 2^n, then a single F(i) for a too large i, is slower than computing F(j) for all j < i.

This is illustrated by the fact that there are more leaves in a complete binary tree than all the other nodes summed.