Answer
See explanation
Work Step by Step
Below is a straightforward algebraic verification that the closed‐form (Binet’s formula) for the Fibonacci sequence satisfies the Fibonacci recurrence. Let
\[
F_n
\;=\; \frac{1}{\sqrt{5}}
\Bigl[\Bigl(\tfrac{1+\sqrt{5}}{2}\Bigr)^{n+1}
-\Bigl(\tfrac{1-\sqrt{5}}{2}\Bigr)^{n+1}\Bigr].
\]
We want to show that this sequence \(F_0, F_1, F_2, \dots\) (as defined by the above expression for each \(n \ge 0\)) satisfies
\[
F_k = F_{k-1} + F_{k-2}
\quad\text{for all integers } k \ge 2.
\]
---
## Step‐by‐Step Verification
Define
\[
\alpha = \frac{1 + \sqrt{5}}{2},
\quad
\beta = \frac{1 - \sqrt{5}}{2}.
\]
Note that both \(\alpha\) and \(\beta\) satisfy the quadratic equation
\[
x^2 = x + 1,
\]
which implies
\[
\alpha^2 = \alpha + 1,
\quad
\beta^2 = \beta + 1.
\]
Hence, the closed‐form formula can be written as
\[
F_n
\;=\; \frac{\alpha^{\,n+1} - \beta^{\,n+1}}{\sqrt{5}}.
\]
### 1. Write Out \(F_k\), \(F_{k-1}\), \(F_{k-2}\)
- \(\displaystyle
F_k
= \frac{\alpha^{k+1} - \beta^{k+1}}{\sqrt{5}},\)
- \(\displaystyle
F_{k-1}
= \frac{\alpha^k - \beta^k}{\sqrt{5}},\)
- \(\displaystyle
F_{k-2}
= \frac{\alpha^{k-1} - \beta^{k-1}}{\sqrt{5}}.
\)
### 2. Check the Recurrence
Compute \(F_{k-1} + F_{k-2}\):
\[
F_{k-1} + F_{k-2}
\;=\;
\frac{\alpha^k - \beta^k}{\sqrt{5}}
\;+\;
\frac{\alpha^{k-1} - \beta^{k-1}}{\sqrt{5}}
\;=\;
\frac{1}{\sqrt{5}}\Bigl[\alpha^k + \alpha^{k-1} - \beta^k - \beta^{k-1}\Bigr].
\]
Group the \(\alpha\)‐terms and the \(\beta\)‐terms:
\[
=\;
\frac{1}{\sqrt{5}}\Bigl[(\alpha^k + \alpha^{k-1}) - (\beta^k + \beta^{k-1})\Bigr].
\]
#### Factor using \(\alpha^2 = \alpha + 1\) and \(\beta^2 = \beta + 1\)
\[
\alpha^k + \alpha^{k-1}
\;=\;
\alpha^{k-1}\,\bigl(\alpha + 1\bigr)
\;=\;
\alpha^{k-1}\,(\alpha^2)
\;=\;
\alpha^{k+1}.
\]
(The last equality holds because \(\alpha^2 = \alpha + 1\).)
Similarly,
\[
\beta^k + \beta^{k-1}
\;=\;
\beta^{k-1}(\beta + 1)
\;=\;
\beta^{k-1}\,\beta^2
\;=\;
\beta^{k+1}.
\]
Therefore,
\[
(\alpha^k + \alpha^{k-1}) - (\beta^k + \beta^{k-1})
\;=\;
\alpha^{k+1} - \beta^{k+1}.
\]
Hence,
\[
F_{k-1} + F_{k-2}
\;=\;
\frac{1}{\sqrt{5}}
\Bigl[\alpha^{k+1} - \beta^{k+1}\Bigr]
\;=\;
F_k.
\]
This shows precisely that
\[
F_k = F_{k-1} + F_{k-2},
\]
as required.
---
## Conclusion
By direct algebraic manipulation and the fact that \(\alpha\) and \(\beta\) are roots of \(x^2 - x - 1 = 0\), we see the closed‐form expression
\[
F_n
= \frac{1}{\sqrt{5}}
\Bigl[\Bigl(\tfrac{1+\sqrt{5}}{2}\Bigr)^{n+1}
- \Bigl(\tfrac{1-\sqrt{5}}{2}\Bigr)^{n+1}\Bigr]
\]
satisfies the Fibonacci recurrence \(F_k = F_{k-1} + F_{k-2}\) for all \(k\ge2\).
