A Fibonacci Sum

Consider the criterion that $\sum_{k=1}^n a_k = F_n$ , then inductively, it better be the case that $F_{n+1} = F_n + a_{n+1} \implies a_{n+1} = F_{n-1}$ from the definition of $F_{n+1}$ . This suggests that for $n > 2$ , we can just set $a_n = F_{n-2}$ (because this is when the recurrence hold); furthermore, we also need to set $a_1 = F_1$ , which leaves exactly one way to set $a_2 = 0$ . Of course, if we use the extended fibonacci sequence, then we can see that $a_n = F_{n-2}$ .

Now, let's look at the summation: $\begin{aligned}\sum_{k=1}^n a_{2k-1} &= a_1 + a_3 + \cdots + a_{2n-1} \\ &= a_1 + \underbrace{(a_1 + a_2)}_{a_3} + \underbrace{(a_3 + a_4)}_{a_5} + \cdots + \underbrace{(a_{2n-3} + a_{2n-2})}_{a_{2n-1}} \\ &= a_1 + \sum_{k=1}^{2n-2} a_k \\ &= a_1 + F_{2n-2}\end{aligned}$

Therefore, $c = a_1 = 1$ and $f(n) = 2n-2$ , which makes $f(2013) + c = 4\boxed{025}$ .

Note that $\sum_{k=1}^n F_{2k-1} = F_{2n}$ . We begin by noting the terms of $a_i$ :

$a_1 = F_1 = 1$

$a_2 = F_2 - a_1 = 0$

$a_3 = F_3 - (a_1 + a_2) = F_3 - F_2 = F_1$

$a_4 = F_4 - (a_3+a_2+a_1) = F_4 - F_3 = F_2$

Hence, $a_{n+2} = F_n$ . Then, we have $\sum_{i=1}^{n} a_{2i-1} = \sum_{k=1}^n F_{2k-3}$ . Because the Fibonacci numbers are (technically) not defined for negative arguments, we rewrite the first sum as $a_1 + \sum_{i=2}^n a_{2i-1}$ in order to rewrite the latter as $1 + \sum_{k=2}^n F_{2k-3} = 1 + F_{2n-2}$ as per the comment made at the beginning of the proof. Hence, $\sum_{i=1}^{n} a_{2i-1} = S_{2n-2} + 1$ , and so $f(2013)+1 = 2(2013)-2 + 1 = 4025$ , whose last three digits are $\boxed{025}$ .