That formula answers the question of "what's the number at the nth position in the series?" To answer the question of "what are the numbers in the nth to the mth position in the series?", we can use the formula (or an adaptation) to get to nth and (n+1)th numbers in the series, and from then on just use addition. Using the formula for each number in the range is inefficient.I've just posted the Binet's formula because it's a curiosity, of course it slow down the PC , but if the user whats a number very big,maybe is it better use the formula? or 0+1+1+2+3+5+8+13+21+... ? .