Question 61
The Fibonacci numbers are defined by $F_{0}=0$, $F_{1}=1$, and $F_{n}=F_{n-1}+F_{n-2}$ for $n \geq 2$. There exist unique positive integers $n_{1}$, $n_{2}$, $n_{3}$, $n_{4}$, $n_{5}$, $n_{6}$ such that
$$
\sum_{i_{1}=0}^{100} \sum_{i_{2}=0}^{100} \sum_{i_{3}=0}^{100} \sum_{i_{4}=0}^{100} \sum_{i_{5}=0}^{100} F_{i_{1}+i_{2}+i_{3}+i_{4}+i_{5}}=F_{n_{1}}-5 F_{n_{2}}+10 F_{n_{3}}-10 F_{n_{4}}+5 F_{n_{5}}-F_{n_{6}}
$$
Find $n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}$.
$$
\sum_{i_{1}=0}^{100} \sum_{i_{2}=0}^{100} \sum_{i_{3}=0}^{100} \sum_{i_{4}=0}^{100} \sum_{i_{5}=0}^{100} F_{i_{1}+i_{2}+i_{3}+i_{4}+i_{5}}=F_{n_{1}}-5 F_{n_{2}}+10 F_{n_{3}}-10 F_{n_{4}}+5 F_{n_{5}}-F_{n_{6}}
$$
Find $n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}$.