1. ## Fibonacci Numbers

Q: Show that the Fibonacci numbers F(1),F(2),F(3),... ,where F(1)= F(2)=1 and F(k)= F(k-2) + F(k-1) for k>2 satisfy the following equality for all n (greater or equal to) 1.

(F(1))^2 + (F(2))^2 + (F(3))^2+.....+ (F(n))^2 = F(n) * F(n+1)

I did the basis step:
1^2 + 1^2 = 1

but how do I do the inductive step for this question?

Thanks,
Creative

2. Originally Posted by Creative
Q: Show that the Fibonacci numbers F(1),F(2),F(3),... ,where F(1)= F(2)=1 and F(k)= F(k-2) + F(k-1) for k>2 satisfy the following equality for all n (greater or equal to) 1.
(F(1))^2 + (F(2))^2 + (F(3))^2+.....+ (F(n))^2 = F(n) * F(n+1)

I did the basis step:
1^2 + 1^2 = 1
You did not do the first step.
$1^2 + 1^2 \ne 1$
Is it true that $1^2 + 1^2 = F(2)F(2+1)?$

3. Oops basis step is 1^2(F1) = 1^2(F2) = 1?

1^2 +1^2 = F(2)*F(2+1)
is not true?
1^2*1^2? = 1

4. Originally Posted by Creative
Oops basis step is 1^2(F1) = 1^2(F2) = 1?
1^2 +1^2 = F(2)*F(2+1) is not true?
Yes it is true.
$1^2 + 1^2=2$
$F(2)=1\;\&\;F(2+1)=F(3)=2$.
$F(2)F(3)=2$

5. Thanks, so basis step would be proving F(1)
F(1)*(F(1+1)) = F(1)*F(2) = (1^2)*(1^2) = 1
so true for basis step.
Then how would I do inductive?

6. I wonder if you understand what the problem really says?
You want to prove $\sum\limits_{k = 1}^n {\left[ {f(k)} \right]^2 } = f(n)f(n + 1)$.
We have more or less done the first step.
Now assume that $\sum\limits_{k = 1}^J {\left[ {f(k)} \right]^2 } = f(J)f(J + 1)$ is true.
On that basis prove it is true for $J+1$.
I am not going to do it for you. For it is your problem.