Let $\alpha$ and $\beta$ be the roots of $x ^ { 2 } - x - 1 = 0$, with $\alpha > \beta$. For all positive integers $n$, define $$\begin{aligned}
& a _ { n } = \frac { \alpha ^ { n } - \beta ^ { n } } { \alpha - \beta } , \quad n \geq 1 \\
& b _ { 1 } = 1 \text { and } \quad b _ { n } = a _ { n - 1 } + a _ { n + 1 } , \quad n \geq 2 .
\end{aligned}$$ Then which of the following options is/are correct? (A) $\quad a _ { 1 } + a _ { 2 } + a _ { 3 } + \cdots + a _ { n } = a _ { n + 2 } - 1$ for all $n \geq 1$ (B) $\quad \sum _ { n = 1 } ^ { \infty } \frac { a _ { n } } { 10 ^ { n } } = \frac { 10 } { 89 }$ (C) $b _ { n } = \alpha ^ { n } + \beta ^ { n }$ for all $n \geq 1$ (D) $\quad \sum _ { n = 1 } ^ { \infty } \frac { b _ { n } } { 10 ^ { n } } = \frac { 8 } { 89 }$
Let $\alpha$ and $\beta$ be the roots of $x ^ { 2 } - x - 1 = 0$, with $\alpha > \beta$. For all positive integers $n$, define
$$\begin{aligned}
& a _ { n } = \frac { \alpha ^ { n } - \beta ^ { n } } { \alpha - \beta } , \quad n \geq 1 \\
& b _ { 1 } = 1 \text { and } \quad b _ { n } = a _ { n - 1 } + a _ { n + 1 } , \quad n \geq 2 .
\end{aligned}$$
Then which of the following options is/are correct?\\
(A) $\quad a _ { 1 } + a _ { 2 } + a _ { 3 } + \cdots + a _ { n } = a _ { n + 2 } - 1$ for all $n \geq 1$\\
(B) $\quad \sum _ { n = 1 } ^ { \infty } \frac { a _ { n } } { 10 ^ { n } } = \frac { 10 } { 89 }$\\
(C) $b _ { n } = \alpha ^ { n } + \beta ^ { n }$ for all $n \geq 1$\\
(D) $\quad \sum _ { n = 1 } ^ { \infty } \frac { b _ { n } } { 10 ^ { n } } = \frac { 8 } { 89 }$