A reader ( Paddy ) surprised me with an excellent answer to the exercise I posted. The series I posted was in fact the running total of the squares of the Fibonacci numbers. Let me put that in a table.
$ \begin{matrix}
n & F(n) & (F(n))^2 & \Sigma\\
1 & 1 & 1 & 1\\
2 & 1 & 1 & 2\\
3 & 2 & 4 & 6\\
4 & 3 & 9 & 15\\
5 & 5 & 25 & 40\\
6 & 8 & 64 & 104\\
7 & 13 & 169 & 273
\end{matrix} $
Paddy's answer...
$ \begin{matrix}
n & F(n) & F(n) * F(n+1)\\
1 & 1 & 1 \\
2 & 1 & 2 \\
3 & 2 & 6 \\
4 & 3 & 15 \\
5 & 5 & 40 \\
6 & 8 & 104 \\
7 & 13 & 273
\end{matrix} $
So $ \sum_{k=1}^{n} F_n^{2} = F_n * F_{n+1} $
An interesting conjecture to prove formally.
2-2024 Quran and mathematics
2 weeks ago
[Sorry about the styling]
ReplyDeleteIf we look at Fn * Fn+1
Fn+1 = Fn + Fn-1
So
Fn * (Fn + Fn-1)
= Fn^2 + Fn * Fn-1
Fn = Fn-1 + Fn-2 so we have
Fn^2 + Fn-1 * (Fn-1 + Fn-2)
= Fn^2 + Fn-1^2 + Fn-1 * Fn-2
etc, etc.
Works?