"Consider the sequence defined by the recurrence relation a(n+2) = a(n+1) + a(n) for n ≥ 1, with initial conditions a(1) = a(2) = 1. This sequence is known as the Fibonacci sequence.
Question: Prove that for any positive integer n, the number a(n)*a(n+2) - (a(n+1))^2 equals (-1)^(n+1)."
Verified answer
Jawaban:
Proof by induction:
Base case:
* For n = 1, we have:
* a(1) = 1
* a(2) = 1
* a(1)*a(3) - (a(2))^2 = 1*1 - 1^2 = 0 = (-1)^(1+1)
Therefore, the base case holds.
Inductive step:
Assume that the statement holds for n = k, i.e.:
```
a(k)*a(k+2) - (a(k+1))^2 = (-1)^(k+1)
```
We want to show that the statement also holds for n = k+1:
```
a(k+1)*a(k+3) - (a(k+2))^2 = (-1)^(k+1+1)
```
Substituting the recurrence relation for a(k+1) and a(k+3), we get:
```
(a(k) + a(k-1)) * (a(k+2) + a(k+1)) - (a(k+1) + a(k))^2
```
Expanding the parentheses and simplifying, we get:
```
a(k) * a(k+2) + a(k-1) * a(k+2) + a(k) * a(k+1) + a(k-1) * a(k+1) - a(k+1)^2 - a(k)^2
```
Combining like terms, we get:
```
a(k) * a(k+2) + (a(k-1) + a(k)) * a(k+1) - a(k+1)^2 - a(k)^2
```
Substituting the inductive hypothesis, we get:
```
(-1)^(k+1) + (a(k-1) + a(k)) * a(k+1) - a(k+1)^2 - a(k)^2
```
Distributing the negative sign, we get:
```
(-1)^(k+1) - (a(k-1) + a(k)) * a(k+1) + a(k+1)^2 + a(k)^2
```
Subtracting (a(k-1) + a(k)) * a(k+1) from both sides, we get:
```
- (a(k-1) + a(k)) * a(k+1) + a(k+1)^2 + a(k)^2 = (-1)^(k+1) - (-1)^(k+1) = 0
```
Therefore, the statement holds for n = k+1.
Conclusion:
By the principle of mathematical induction, the statement holds for all positive integers n.
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