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## Tuesday, May 4, 2010

### Exercise - comments

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.

$\begin{matrix} n & F(n) & (F(n))^2 & \Sigma\\ 1 & 1 & 1 & 1\\ 2 & 1 & 1 & 2\\ 3 & 2 & 4 & 6\\ 4 & 3 & 9 & 15\\ 5 & 5 & 25 & 40\\ 6 & 8 & 64 & 104\\ 7 & 13 & 169 & 273 \end{matrix}$

$\begin{matrix} n & F(n) & F(n) * F(n+1)\\ 1 & 1 & 1 \\ 2 & 1 & 2 \\ 3 & 2 & 6 \\ 4 & 3 & 15 \\ 5 & 5 & 40 \\ 6 & 8 & 104 \\ 7 & 13 & 273 \end{matrix}$

So $\sum_{k=1}^{n} F_n^{2} = F_n * F_{n+1}$
An interesting conjecture to prove formally.

#### 1 comment:

1. [Sorry about the styling]
If 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?

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