For any integer $n \geq 1$, define $a_n = \frac{1000^n}{n!}$. Then the sequence $\{a_n\}$
(A) does not have a maximum
(B) attains maximum at exactly one value of $n$
(C) attains maximum at exactly two values of $n$
(D) attains maximum for infinitely many values of $n$

For any integer $n \geq 1$, define $a _ { n } = \frac { 1000 ^ { n } } { n ! }$. Then the sequence $\left\{ a _ { n } \right\}$
(A) does not have a maximum
(B) attains maximum at exactly one value of $n$
(C) attains maximum at exactly two values of $n$
(D) attains maximum for infinitely many values of $n$

The limit $$\lim _ { n \rightarrow \infty } \left( 2 ^ { - 2 ^ { n + 1 } } + 2 ^ { - 2 ^ { n - 1 } } \right) ^ { 2 ^ { - n } }$$ equals
(A) 1.
(B) $\frac { 1 } { \sqrt { 2 } }$.
(C) 0.
(D) $\frac { 1 } { 4 }$.

Let $p , q$ be integers and let $\alpha , \beta$ be the roots of the equation, $x ^ { 2 } - x - 1 = 0$, where $\alpha \neq \beta$. For $n = 0,1,2 , \ldots$, let $a _ { n } = p \alpha ^ { n } + q \beta ^ { n }$.
FACT: If $a$ and $b$ are rational numbers and $a + b \sqrt { 5 } = 0$, then $a = 0 = b$.
$a _ { 12 } =$
[A] $a _ { 11 } - a _ { 10 }$
[B] $a _ { 11 } + a _ { 10 }$
[C] $2 a _ { 11 } + a _ { 10 }$
[D] $a _ { 11 } + 2 a _ { 10 }$

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 $\alpha , \beta$ be the distinct roots of the equation $x ^ { 2 } - \left( t ^ { 2 } - 5 t + 6 \right) x + 1 = 0 , t \in \mathbb { R }$ and $a _ { n } = \alpha ^ { n } + \beta ^ { n }$. Then the minimum value of $\frac { a _ { 2023 } + a _ { 2025 } } { a _ { 2024 } }$ is
(1) $- 1 / 4$
(2) $- 1 / 4$
(3) $- 1 / 2$
(4) $1 / 4$

If $\alpha , \beta$ are the roots of the equation, $\mathrm { x } ^ { 2 } - \mathrm { x } - 1 = 0$ and $\mathrm { S } _ { \mathrm { n } } = 2023 \alpha ^ { \mathrm { n } } + 2024 \beta ^ { \mathrm { n } }$, then
(1) $2 \quad \mathrm {~S} _ { 12 } = \mathrm { S } _ { 11 } + \mathrm { S } _ { 10 }$
(2) $\mathrm { S } _ { 12 } = \mathrm { S } _ { 11 } + \mathrm { S } _ { 10 }$
(3) $2 \mathrm {~S} _ { 11 } = \mathrm { S } _ { 12 } + \mathrm { S } _ { 10 }$
(4) $\mathrm { S } _ { 11 } = \mathrm { S } _ { 10 } + \mathrm { S } _ { 12 }$

Let $\alpha , \beta ; \alpha > \beta$, be the roots of the equation $x ^ { 2 } - \sqrt { 2 } x - \sqrt { 3 } = 0$. Let $\mathrm { P } _ { n } = \alpha ^ { n } - \beta ^ { n } , n \in \mathrm {~N}$. Then $( 11 \sqrt { 3 } - 10 \sqrt { 2 } ) \mathrm { P } _ { 10 } + ( 11 \sqrt { 2 } + 10 ) \mathrm { P } _ { 11 } - 11 \mathrm { P } _ { 12 }$ is equal to
(1) $10 \sqrt { 3 } \mathrm { P } _ { 9 }$
(2) $11 \sqrt { 3 } P _ { 9 }$
(3) $10 \sqrt { 2 } \mathrm { P } _ { 9 }$
(4) $11 \sqrt { 2 } \mathrm { P } _ { 9 }$

Given a real number sequence $\left\langle a _ { n } \right\rangle$ satisfying $a _ { 1 } = 1 , a _ { n + 1 } = \frac { 2 n + 1 } { 2 n - 1 } a _ { n } , n$ is a positive integer. Select the correct options.
(1) $a _ { 2 } = 3$
(2) $a _ { 4 } = 9$
(3) $\left\langle a _ { n } \right\rangle$ is a geometric sequence
(4) $\sum _ { n = 1 } ^ { 20 } a _ { n } = 400$
(5) $\lim _ { n \rightarrow \infty } \frac { a _ { n } } { n } = 2$