# Fibonacci sequence proof

• May 20th 2010, 06:36 AM
mattgavin
Fibonacci sequence proof
Hi,

I'm having trouble solving the following proof on my study guide for an exam:

Prove for the Fibonacci sequence F_k

F_k*F_k - F_(k-1)*F_(k-1) = F_k*F_(k+1) - F_(k+1)*F_(k-1)

for k>=1

Could anyone show me what I need to be doing? Any help would be appreciated. Thanks!
• May 20th 2010, 06:57 AM
alexmahone
$F_k+F_{k-1}=F_{k+1}$

Multiplying both sides of the equation by $F_k-F_{k-1}$,

$(F_k+F_{k-1})(F_k-F_{k-1})=F_{k+1}(F_k-F_{k-1})$

$F_k^2-F_{k-1}^2=F_kF_{k+1}-F_{k+1}F_{k-1}$

$F_kF_k-F_{k-1}F_{k-1}=F_kF_{k+1}-F_{k+1}F_{k-1}$
• May 20th 2010, 07:16 AM
Quote:

Originally Posted by mattgavin
Hi,

I'm having trouble solving the following proof on my study guide for an exam:

Prove for the Fibonacci sequence F_k

F_k*F_k - F_(k-1)*F_(k-1) = F_k*F_(k+1) - F_(k+1)*F_(k-1)

for k>=1

Could anyone show me what I need to be doing? Any help would be appreciated. Thanks!

Alternatively, the Fibonacci recursion relation is

$F_{k+1}=F_k+F_{k-1}$

We are asked to prove $F_kF_k-F_{k-1}F_{k-1}=F_kF_{k+1}-F_{k+1}F_{k-1}$

This contains $F_{k-1}$ on the left, hence $F_{k-1}=F_{k+1}-F_k$

Therefore, rewriting the left side

$F_kF_k-F_{k-1}F_{k-1}=F_kF_k-\left(F_{k+1}-F_k\right)\left(F_{k+1}-F_k\right)$

$=F_kF_k-\left(F_{k+1}F_{k+1}-2F_{k+1}F_k+F_kF_k\right)$

$=2F_kF_{k+1}-F_{k+1}F_{k+1}=F_kF_{k+1}+F_kF_{k+1}-F_{k+1}F_{k+1}$

$F_{k+1}$ is a factor of the last 2 terms (it's a factor of all 3 of course), giving

$F_kF_{k+1}+F_{k+1}\left(F_k-F_{k+1}\right)=F_kF_{k+1}-F_{k+1}\left(F_{k+1}-F_k\right)$

In brackets is $F_{k-1}$
• May 20th 2010, 07:28 AM
wonderboy1953
Letting you know
Just picked up a book on Fibonacci numbers by Alfred Posamentier (2007) which is interesting reading.