Answer
See explanation
Work Step by Step
## **Statement to Prove**
For all integers \(n \ge 0\),
\[
F_{n+2}\,F_{n} \;-\; F_{n+1}^{2} \;=\; (-1)^{\,n+1}.
\]
---
## **Proof by Mathematical Induction**
### 1. Base Case \((n=0)\)
We check the identity for \(n=0\). The left‐hand side (LHS) is
\[
F_{2}\,F_{0} \;-\; F_{1}^{2}.
\]
Using \(F_{0}=0, F_{1}=1, F_{2}=1\):
\[
\text{LHS} = (1)\cdot(0) \;-\; 1^{2} = 0 \;-\; 1 = -1.
\]
Meanwhile, the right‐hand side (RHS) is
\[
(-1)^{\,0+1} = (-1)^{1} = -1.
\]
Hence, for \(n=0\) both sides match (\(-1 = -1\)), so the base case holds.
---
### 2. Inductive Step
Assume the statement is true for some integer \(n\). In other words, **inductive hypothesis**:
\[
F_{n+2}\,F_{n} \;-\; F_{n+1}^{2} = (-1)^{\,n+1}.
\]
We want to prove it then holds for \((n+1)\). That is, we want to show
\[
F_{(n+1)+2}\,F_{n+1} \;-\; F_{(n+1)+1}^{2}
\;=\;
F_{n+3}\,F_{n+1} \;-\; F_{n+2}^{2}
\;=\;
(-1)^{\,(n+1)+1} \;=\; (-1)^{\,n+2}.
\]
#### Algebraic Manipulation
Start with \(F_{n+3} = F_{n+2} + F_{n+1}\) (by the Fibonacci recurrence). Then
\[
\begin{aligned}
F_{n+3}\,F_{n+1} \;-\; F_{n+2}^{2}
&=\;
\bigl(F_{n+2} + F_{n+1}\bigr)\,F_{n+1}
\;-\;
F_{n+2}^{2}
\\[6pt]
&=\;
F_{n+2}\,F_{n+1} \;+\; F_{n+1}^{2}
\;-\;
F_{n+2}^{2}.
\end{aligned}
\]
Re‐group terms to compare with the inductive hypothesis \(\bigl(F_{n+2}F_{n} - F_{n+1}^{2}\bigr)\). Notice:
\[
F_{n+2}\,F_{n+1}
\;-\;
F_{n+2}^{2}
\;=\;
F_{n+2}\,\bigl(F_{n+1} - F_{n+2}\bigr)
\;=\;
-\,F_{n+2}\,\bigl(F_{n+2} - F_{n+1}\bigr).
\]
But \(F_{n+2} - F_{n+1} = F_{n}\) (since \(F_{n+2} = F_{n+1} + F_{n}\) rearranged). Hence
\[
F_{n+2}\,F_{n+1} - F_{n+2}^{2}
\;=\;
-\,F_{n+2}\,\bigl(F_{n+2} - F_{n+1}\bigr)
\;=\;
-\,F_{n+2}\,F_{n}.
\]
Thus
\[
F_{n+3}\,F_{n+1} \;-\; F_{n+2}^{2}
\;=\;
\bigl(-\,F_{n+2}\,F_{n}\bigr)
\;+\;
F_{n+1}^{2}
\;=\;
-\bigl(F_{n+2}\,F_{n} - F_{n+1}^{2}\bigr).
\]
But by the inductive hypothesis,
\[
F_{n+2}\,F_{n} - F_{n+1}^{2} = (-1)^{\,n+1}.
\]
Therefore,
\[
F_{n+3}\,F_{n+1} - F_{n+2}^{2}
\;=\;
-\bigl[\,(-1)^{\,n+1}\bigr]
\;=\;
(-1)^{\,n+2}.
\]
This completes the induction step, proving the identity for \(n+1\).
---
### Conclusion
By mathematical induction, for all integers \(n\ge 0\),
\[
\boxed{F_{n+2}\,F_{n} \;-\; F_{n+1}^{2} \;=\; (-1)^{\,n+1}.}
\]
