A ratio of Fibonacci sums

Let FnF_n be the nn-th Fibonacci number (F1=F2=1F_1 = F_2 = 1). Compute F1+F2+F3+⋯+F300F3+F6+F9+⋯+F300\dfrac{F_1 + F_2 + F_3 + \dots + F_{300}}{F_3 + F_6 + F_9 + \dots + F_{300}}.

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  1. ∑k=1nFk=Fn+2−1\sum_{k=1}^{n}F_k=F_{n+2}-1 and ∑j=1mF3j=F3m+2−12\sum_{j=1}^{m}F_{3j}=\tfrac{F_{3m+2}-1}{2}.
  2. Both involve F302−1F_{302}-1; take the ratio.

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