Let u_n = frac{1}{sqrt{5}} left[ left( frac{1 | vedclass

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(A) The given expression is the Binet's formula for the $n$-th Fibonacci number,where $u_n = F_n$.Let $\alpha = \frac{1 + \sqrt{5}}{2}$ and $\beta = \

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$u_{n+1} = u_n + u_{n-1}$

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(A) The given expression is the Binet's formula for the $n$-th Fibonacci number,where $u_n = F_n$.Let $\alpha = \frac{1 + \sqrt{5}}{2}$ and $\beta = \frac{1 - \sqrt{5}}{2}$.Then $u_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}$.We know that $\alpha$ and $\beta$ are roots of the quadratic equation $x^2 - x - 1 = 0$,so $\alpha^2 = \alpha + 1$ and $\beta^2 = \beta + 1$.For $n \ge 1$,$u_{n+1} = \frac{\alpha^{n+1} - \beta^{n+1}}{\sqrt{5}}$.$u_n + u_{n-1} = \frac{1}{\sqrt{5}} [(\alpha^n - \beta^n) + (\alpha^{n-1} - \beta^{n-1})] = \frac{1}{\sqrt{5}} [\alpha^{n-1}(\alpha + 1) - \beta^{n-1}(\beta + 1)]$.Since $\alpha + 1 = \alpha^2$ and $\beta + 1 = \beta^2$,we get $u_n + u_{n-1} = \frac{\alpha^{n+1} - \beta^{n+1}}{\sqrt{5}} = u_{n+1}$.Thus,the recurrence relation is $u_{n+1} = u_n + u_{n-1}$.

Let $u_n = \frac{1}{\sqrt{5}} \left[ \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n \right]$,for $n = 0, 1, 2, ...$. Then which of the following is true?

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