The sum of Fibonacci sequence

Let $S_n = \displaystyle \sum_{k=0}^n F_k$ . Then we have $S_0 = 0 = F_2 - 1$ , $S_1 = 1 = F_3 - 1$ , $S_2 = 2 = F_4 - 1$ , $S_3 = 4 = F_5 -1$ , $\cdots$ . It would appear that $S_n = F_{n+2}-1$ . We can prove the claim is true for all $n \ge 0$ by induction . The claim is true for $n=0$ . Assuming the claim is true for $n$ , then

$S_{n+1} = S_n + F_{n+1} = F_{n+2} - 1 + F_{n+1} = F_{n+3} - 1$

The claim is also true for $n+1$ , therefore $\displaystyle \sum_{k=0}^n F_k = \boxed{F_{n+2}-1}$ is true for all $n \ge 0$ .

Let $S=F_{1}+F_{2}+F_{3}+ \cdots +F_{n}. \newline$ So $\newline 2S \newline =(F_{1}+F_{1})+(F_{2}+F_{2})+(F_{3}+ \cdots F_{n-2}) + (F_{n-1} + F_{n-1}) + (F_{n} + F_{n}) \newline =F_{1}+(F_{1}+F_{2})+(F_{2}+F_{3})+ \cdots (F_{n-2} + F_{n-1}) + (F_{n-1} + F_{n}) + F_{n}. \newline$ And we also know: $F_{n} = F_{n-1} + F_{n-2}. \newline$ So $\newline 2S \newline = F_{1} + F_{3} + F_{4} + \cdots + F_{n+1} + F_{n} \newline = S - F_{2} + (F_{n+1} + F_{n}) \newline 2S = S - 1 + F_{n+2} \newline \boxed{S = F_{n+2} - 1}$