Let $\alpha$ and $\beta$ be the roots of $x^{2} - x - 1 = 0$, with $\alpha > \beta$. For all positive integers n, define $a_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}$, $n \ge 1$, $b_1 = 1$ and $b_n = a_{n-1} + a_{n+1}$, $n \ge 2$ Then which of the following options is/are correct ?

Let α and β be the roots of x² - x - 1 = 0, with α > β. For all positive integers n, define aₙ = (αⁿ - βⁿ)/(α - β), n ≥ 1, b₁ = 1 and bₙ = aₙ₋₁ + aₙ₊₁, n ≥ 2 Then which of the following options is/are correct ?

$\alpha, \beta \text{ roots of } x^{2} - x - 1 = 0 \Rightarrow \alpha + \beta = 1,\ \alpha\beta = -1$

$a_n = \dfrac{\alpha^n - \beta^n}{\alpha - \beta}$

$b_1 = 1,\quad b_n = a_{n-1} + a_{n+1}\ (n \ge 2)$

$\text{Check: } \sum_{n=1}^{\infty} \dfrac{a_n}{10^n} = \dfrac{10}{89}$

$b_n = \alpha^n + \beta^n\ ?\qquad a_1 + \cdots + a_n = a_{n+2} - 1\ ?$

$\sum_{n=1}^{\infty} \dfrac{b_n}{10^n} = \dfrac{8}{89}\ ?$

$\text{1. Show } a_n = F_n\ (\text{Fibonacci})$

$\text{2. Verify } (1), (2), (3) \text{, disprove } (4)$

$a_1 = 1,\quad a_2 = \alpha + \beta = 1$

$\alpha^2 = \alpha+1 \Rightarrow \alpha^{n} = \alpha^{n-1} + \alpha^{n-2}$

$\therefore\ a_n = a_{n-1} + a_{n-2}\quad (a_n = F_n)$

$a_k = a_{k+2} - a_{k+1}$

$\sum_{k=1}^{n} a_k = a_{n+2} - a_2 = a_{n+2} - 1\ \checkmark$

$b_n = a_{n-1} + a_{n+1} = \dfrac{(\alpha^{n-1}+\alpha^{n+1}) - (\beta^{n-1}+\beta^{n+1})}{\alpha-\beta}$

$= \alpha^{n} + \beta^{n}\quad (\text{using } \alpha\beta = -1)\ \checkmark$

$b_1 = 1 = \alpha+\beta,\qquad b_2 = a_1 + a_3 = 3 = \alpha^2+\beta^2$

$b_3 = a_2 + a_4 = 4$

$S = \sum_{n=1}^{\infty} a_n x^{n} = x + xS + x^{2}S$

$S = \dfrac{x}{1 - x - x^{2}}$

$S = \dfrac{1/10}{1 - 1/10 - 1/100} = \dfrac{1/10}{89/100}$

$\sum_{n=1}^{\infty} \dfrac{a_n}{10^n} = \dfrac{10}{89}\ \checkmark$

$\sum_{n=1}^{\infty} \dfrac{b_n}{10^n} = \dfrac{\alpha}{10-\alpha} + \dfrac{\beta}{10-\beta}$

$= \dfrac{10(\alpha+\beta) - 2\alpha\beta}{100 - 10(\alpha+\beta) + \alpha\beta} = \dfrac{12}{89}$

$\dfrac{12}{89} \ne \dfrac{8}{89}\ \Rightarrow\ (4)\ \text{is false}$

$\text{Correct options: } (1),\ (2),\ (3)$