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After some tedious exploration:
Denote k 0 = 0 and k i = k i − 1 + F 2 i for all positive integers i , (i.e., the sequence is 1, 4, 12, 33, ...) then
1 + F 2 n − 1 2 = ( k n − 1 k n − 1 + 2 ) ( k n + 2 k n ) .
Now if we express each term 1 + F 2 n − 1 2 in the form above, it telescopes (the terms cancel out) leaving only the first term, and since k n + 2 k n tends to 1 as n becomes huge, the product will converge to k 1 k 1 + 2 = 3 .
(As for why it works, consider the expansion of F n = 5 1 ( ( 2 1 + 5 ) n − ( 2 1 − 5 ) n ) . Using the identity 1 + x + x 2 + . . . + x n = x − 1 x n + 1 − 1 we can express k n in terms of powers of 2 1 + 5 and 2 1 − 5 . Expansion will give you the first equation.)