## Sum of first n Fibonacci numbers with respect to n?

I know that the nth Fibonacci number is defined as:

$$\frac{{1+\sqrt{5}}^{n}-{1-\sqrt{5}}^{n}}{{2}^{n}\sqrt{5}}$$

But may I know the formula for the sum of the first n Fibonacci numbers with respect to n? Thanks.

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 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Hi dimension10! That formula you give can't possibly be correct, since it evaluates to 0... Did you forget to add some brackets? Anyway, the most elegant formula for the sum of the first n Fibonacci numbers is $$F_0+F_1+...+F_n=F_{n+2}-1$$ Using the (correct) formula for $F_{n+2}$ gives you the desired formula. Check http://en.wikipedia.org/wiki/Fibonacci_number
 Admin Isn't it just an obvious application of $$F_n = F_{n-1} + F_{n-2}$$ definition?

## Sum of first n Fibonacci numbers with respect to n?

 Quote by micromass Hi dimension10! That formula you give can't possibly be correct, since it evaluates to 0... Did you forget to add some brackets?
Yes.I meant
$$\frac{{(1+\sqrt{5})}^{n}-{(1-\sqrt{5})}^{n}}{{2}^{n}\sqrt{5}}$$

 Quote by micromass Anyway, the most elegant formula for the sum of the first n Fibonacci numbers is $$F_0+F_1+...+F_n=F_{n+2}-1$$ Using the (correct) formula for $F_{n+2}$ gives you the desired formula. Check http://en.wikipedia.org/wiki/Fibonacci_number
Thanks.

 So we could write it as: $$\frac{{(1+\sqrt{5})}^{n+2}-{(1-\sqrt{5})}^{n+2}}{{2}^{n+2}\sqrt{5}}-1$$