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.

13-2016 Open letter to Open Source for You (OSFY)

6 months 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?