Question

Prove that ${\mathbf{f}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{+}{\mathbf{f}}_{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{-}{\mathbf{f}}_{\mathbf{n}}^{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{n}}$where n is a positive integer.

Short answer

${\mathrm{f}}_{\mathrm{n}+1}+{\mathrm{f}}_{\mathrm{n}-1}-{\mathrm{f}}_{\mathrm{n}}^2=(-1{)}^{\mathrm{n}}$whenever n is a positive integer

Step-by-step solution

 Step 1: The definition of Fibonacci numbers

By the definition of Fibonacci numbers

$\begin{array}{r}f\left(0\right)=0\\ f\left(1\right)=1\\ f_n=f_{n-1}+f_{n-2}\text{when}n\ge 2\end{array}$

To prove that fn+1+fn−1−fn2=(−1)n whenever n is a positive integer

The proof is given by INDUCTION:

Let P(n) be $f_{n+1}+f_{n-1}-f_n^2=(-1{)}^n$

Inductive step: Assume that P(k) is true.

To prove that P(K+1) is also true.

Therefore, it can be written as:

$\begin{array}{r}{\mathrm{f}}_{(\mathrm{k}+1)+1}+{\mathrm{f}}_{(\mathrm{k}+1)-1}-{\mathrm{f}}_{\mathrm{k}+1}^2={\mathrm{f}}_{\mathrm{k}+2}{\mathrm{f}}_{\mathrm{k}}-{\mathrm{f}}_{\mathrm{k}+1}^2\\ =\left({\mathrm{f}}_{\mathrm{k}}+{\mathrm{f}}_{\mathrm{k}+1}\right){\mathrm{f}}_{\mathrm{k}}-{\mathrm{f}}_{\mathrm{k}+1}^2\\ ={\mathrm{f}}_2^{\mathrm{k}}+{\mathrm{f}}_{\mathrm{k}+1}{\mathrm{f}}_{\mathrm{k}}-{\mathrm{f}}_{\mathrm{k}+1}^2\\ ={\mathrm{f}}_2^{\mathrm{k}}+{\mathrm{f}}_{\mathrm{k}+1}\left({\mathrm{f}}_{\mathrm{k}+1}-{\mathrm{f}}_{\mathrm{k}})\right.\end{array}$

Now, by the definaton $f_{n-2}=f_n-f_{n-1}$

$\begin{array}{r}{f}_{(k+1)+1}+f_{(k+1)-1}-f_{k+1}^2=f_k^2-f_{k+1}{f}_{k-1}\\ =-\left(f_{k+1}{f}_{k-1}-f_k^2\right)\\ =-(-1{)}^k\\ =(-1{)}^{k+1}\end{array}$[Since P(k) is true]

Hence, it is proved that P (k + 1) is true.

Thus, it is proved by the principle of mathematical induction that:

$f_{n+1}+f_{n-1}-f_n^2=(-1{)}^n$whenever n is a positive integer.