Prove the following identity:

(F(n+1))^2 - (F(n))^2 = F(n-1)F(n+2)

I am trying to do this with induction, and I am stuck!

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- Feb 4th 2011, 04:58 PMveronicak5678Fibonacci Proof
Prove the following identity:

(F(n+1))^2 - (F(n))^2 = F(n-1)F(n+2)

I am trying to do this with induction, and I am stuck! - Feb 4th 2011, 11:15 PMDrexel28
- Feb 5th 2011, 02:47 AMArchie Meade
Alternatively (using induction to practice! given the clarity of Drexel's response) if

$\displaystyle \left(F_{n+1}\right)^2-\left(F_n\right)^2=F_{n-1}F_{n+2}$

then we require that

$\displaystyle \left(F_{n+2}\right)^2-\left(F_{n+1}\right)^2=F_nF_{n+3}$

**Proof**

$\displaystyle F_nF_{n+3}=F_n\left(F_{n+1}+F_{n+2}\right)=F_nF_{n +1}+F_nF_{n+2}$

Therefore, is

$\displaystyle \left(F_{n+2}\right)^2-\left(F_{n+1}\right)^2=F_nF_{n+1}+F_nF_{n+2}\;\;?$

$\displaystyle \left(F_{n+2}\right)^2-F_nF_{n+2}=\left(F_{n+1}\right)^2+F_nF_{n+1}\;\;?$

$\displaystyle F_{n+2}\left[F_{n+2}-F_n\right]=F_{n+1}\left[F_{n+1}+F_n\right]\;\;?$

$\displaystyle F_{n+2}=F_{n+1}+F_n\Rightarrow\ F_{n+2}-F_n=F_{n+1}$

gives

$\displaystyle F_{n+2}F_{n+1}=F_{n+1}F_{n+2}\;\;?$

hence the hypothesis is true