Large Fibonacci Numbers Part 2

amazing part of mathematics.

We will show, by induction on $n$ , that $F_{2n}=\sum_{k=1}^nF_{2k-1}$ .

For $n=1$ we have $F_2=F_1=1$ .

Now $F_{2(n+1)}=F_{2n+1}+F_{2n}=F_{2n+1}+\sum_{k=1}^nF_{2k-1}=\sum_{k=1}^{n+1}F_{2k-1}$ .

For $n=20$ we have $\sum_{k=1}^{20}F_{2k-1}=F_{40}=\boxed{102334155}$ .

In Fibonacci sequence, $F_{n+2}=F_{n+1}+F_{n}$

Therefore we can write $F_{40}=F_{39}+F_{38}$

$F_{38}=F_{37}+F_{36}$

$F_{36}=F_{35}+F_{34} .................$

proceeding this way, and expanding all even terms, only the odd terms remain.

Therefore, $F_{40}=F_{39}+F_{37}+F_{35}+ ...... +F_{1}.$

Hence we get the sum given to be equal to $F_{40}.$

Notice F1+F3 = F4, F1 + F3 + F5 =F6 therefore conclude F1 + F3 + ...+F39 = F40