How to prove this equation by induction?

Question

I am trying to prove this equation by mathematical induction $$f_{n+1}f_{n-1} = f_{n}^{2}+(-1)^n$$ is true where $f_{n} = $ the nth number in the Fibonacci sequence. I don't quite get how to do this for the case of $n+1$. Can anyone provide a step-by-step solution?

Thanks.

Answer 1

Hint: 1. Check $n=1$ and $n=2$ 2. Assume it's true for $n$ and $n+1$, check the case of $n+2$

Answer 2

Assume the statement is true for $n$: $f_{n+2}f_n = (f_{n+1}+f_n)(f_{n+1}-f_{n-1}) = f^2_{n+1} - f_{n+1}f_{n-1} + f^2_n = f^2_{n+1} +(-1)(-1)^n = f^2_{n+1} + (-1)^{n+1}$

Answer 3

Assume $f_{n+1}\ f_{n-1} = f_{n}^2+(-1)^{n}$. Then

$$\begin{align*} f_{n+2}\ f_{n} &= (f_{n+1}+f_{n})\ f_{n} &&\text{Definition of }f_{n+2}\\ &= f_{n+1}\ f_{n}+f_{n}^2&&\text{Distributive property}\\ &= f_{n+1}\ f_{n} + \left[f_{n+1}\ f_{n-1}-(-1)^n\right]&&\text{Induction assumption}\\ &= f_{n+1}\left[f_{n} + f_{n-1}\right]-(-1)^{n}&&\text{Factorisation}\\ &= f_{n+1}^2+(-1)^{n+1}&&\text{Definition of }f_{n+1}\end{align*}$$

Answer 4

Hint:-

$$\begin{pmatrix}f_{n+1} & f_n\\f_n& f_{n-1}\end{pmatrix}=\begin{pmatrix}1 & 1\\1&0\end{pmatrix}^n$$