[Discrete Mathematics] A sequence $\left\{ a _ { n } \right\}$ satisfies $$\left\{ \begin{array} { l } a _ { 1 } = 2 , a _ { 2 } = 5 \\ a _ { n } = 2 a _ { n - 1 } + a _ { n - 2 } \end{array} \quad ( n \geqq 3 ) \right.$$ What is the value of $a _ { 5 }$? [3 points]
(1) 70
(2) 72
(3) 74
(4) 76
(5) 78

When $\lim _ { n \rightarrow \infty } \frac { a \times 6 ^ { n + 1 } - 5 ^ { n } } { 6 ^ { n } + 5 ^ { n } } = 4$, what is the value of the constant $a$? [2 points]
(1) $\frac { 1 } { 3 }$
(2) $\frac { 1 } { 2 }$
(3) $\frac { 2 } { 3 }$
(4) $\frac { 4 } { 3 }$
(5) $\frac { 3 } { 2 }$

The sequence $\left\{ a _ { n } \right\}$ satisfies $a _ { 1 } = 1$ and $$a _ { n + 1 } = n + 1 + \frac { ( n - 1 ) ! } { a _ { 1 } a _ { 2 } \cdots a _ { n } } \quad ( n \geqq 1 )$$ The following is part of the process of finding the general term $a _ { n }$.
For all natural numbers $n$, $$a _ { 1 } a _ { 2 } \cdots a _ { n } a _ { n + 1 } = a _ { 1 } a _ { 2 } \cdots a _ { n } \times ( n + 1 ) + ( n - 1 ) !$$ If $b _ { n } = \frac { a _ { 1 } a _ { 2 } \cdots a _ { n } } { n ! }$, then $b _ { 1 } = 1$ and $$b _ { n + 1 } = b _ { n } + ( \text{(a)} )$$ The general term of the sequence $\left\{ b _ { n } \right\}$ is $b _ { n } =$ (b) so $\frac { a _ { 1 } a _ { 2 } \cdots a _ { n } } { n ! } =$ (b). $\vdots$ Therefore, $a _ { 1 } = 1$ and $a _ { n } = \frac { ( n - 1 ) ( 2 n - 1 ) } { 2 n - 3 }$ for $n \geqq 2$.
When the expression that fits (a) is $f ( n )$ and the expression that fits (b) is $g ( n )$, what is the value of $f ( 13 ) \times g ( 7 )$? [4 points]
(1) $\frac { 1 } { 70 }$
(2) $\frac { 1 } { 77 }$
(3) $\frac { 1 } { 84 }$
(4) $\frac { 1 } { 91 }$
(5) $\frac { 1 } { 98 }$

The sequence $\left\{ a _ { n } \right\}$ satisfies $a _ { 1 } = 1$ and $$a _ { n + 1 } = n + 1 + \frac { ( n - 1 ) ! } { a _ { 1 } a _ { 2 } \cdots a _ { n } } \quad ( n \geqq 1 )$$ The following is part of the process of finding the general term $a _ { n }$.
For all natural numbers $n$, $$a _ { 1 } a _ { 2 } \cdots a _ { n } a _ { n + 1 } = a _ { 1 } a _ { 2 } \cdots a _ { n } \times ( n + 1 ) + ( n - 1 ) !$$ If $b _ { n } = \frac { a _ { 1 } a _ { 2 } \cdots a _ { n } } { n ! }$, then $b _ { 1 } = 1$ and $$b _ { n + 1 } = b _ { n } + ( \text{(가)} )$$ The general term of the sequence $\left\{ b _ { n } \right\}$ is $b _ { n } =$ (나) so $\frac { a _ { 1 } a _ { 2 } \cdots a _ { n } } { n ! } =$ (나). $\vdots$ Therefore, $a _ { 1 } = 1$ and $a _ { n } = \frac { ( n - 1 ) ( 2 n - 1 ) } { 2 n - 3 } ( n \geqq 2 )$.
When the expression that fits (가) is $f ( n )$ and the expression that fits (나) is $g ( n )$, what is the value of $f ( 13 ) \times g ( 7 )$? [4 points]
(1) $\frac { 1 } { 70 }$
(2) $\frac { 1 } { 77 }$
(3) $\frac { 1 } { 84 }$
(4) $\frac { 1 } { 91 }$
(5) $\frac { 1 } { 98 }$

The sequence $\{a_n\}$ satisfies the following for all natural numbers $n$: $$\sum_{k=1}^{n} a_k = \log \frac{(n+1)(n+2)}{2}$$ Let $\sum_{k=1}^{20} a_{2k} = p$. Find the value of $10^p$. [4 points]

What is the value of $\lim _ { n \rightarrow \infty } \frac { 5 ^ { n + 1 } + 2 } { 5 ^ { n } + 3 ^ { n } }$? [2 points]
(1) 2
(2) 3
(3) 4
(4) 5
(5) 6

For a sequence $\left\{ a _ { n } \right\}$ with $a _ { 1 } = 1$ and satisfying
$$a _ { n + 1 } = \frac { 2 n } { n + 1 } a _ { n }$$
for all natural numbers $n$, what is the value of $a _ { 4 }$? [3 points]
(1) $\frac { 3 } { 2 }$
(2) 2
(3) $\frac { 5 } { 2 }$
(4) 3
(5) $\frac { 7 } { 2 }$

For a sequence $\left\{ a _ { n } \right\}$ with first term 1, let $S _ { n } = \sum _ { k = 1 } ^ { n } a _ { k }$. The following holds:
$$n S _ { n + 1 } = ( n + 2 ) S _ { n } + ( n + 1 ) ^ { 3 } \quad ( n \geq 1 )$$
The following is part of the process of finding the general term of the sequence $\left\{ a _ { n } \right\}$.
Since $S _ { n + 1 } = S _ { n } + a _ { n + 1 }$ for natural numbers $n$,
$$n a _ { n + 1 } = 2 S _ { n } + ( n + 1 ) ^ { 3 } \quad \cdots (\text{ㄱ})$$
For natural numbers $n \geq 2$,
$$( n - 1 ) a _ { n } = 2 S _ { n - 1 } + n ^ { 3 } \quad \cdots (\text{ㄴ})$$
Subtracting (ㄴ) from (ㄱ), we obtain
$$n a _ { n + 1 } = ( n + 1 ) a _ { n } + \text{ (A) }$$
Dividing both sides by $n ( n + 1 )$,
$$\frac { a _ { n + 1 } } { n + 1 } = \frac { a _ { n } } { n } + \frac { \text{ (A) } } { n ( n + 1 ) }$$
Let $b _ { n } = \frac { a _ { n } } { n }$. Then
$$b _ { n + 1 } = b _ { n } + 3 + \text{ (B) } \quad ( n \geq 2 )$$
Therefore
$$b _ { n } = b _ { 2 } + \text{ (C) } \quad ( n \geq 3 )$$
holds.
What are the correct expressions for (A), (B), and (C)?

For a sequence $\left\{ a _ { n } \right\}$ with first term 1, let $S _ { n } = \sum _ { k = 1 } ^ { n } a _ { k }$. The following relation holds:
$$n S _ { n + 1 } = ( n + 2 ) S _ { n } + ( n + 1 ) ^ { 3 } \quad ( n \geq 1 )$$
The following is part of the process of finding the general term of the sequence $\left\{ a _ { n } \right\}$.
Since $S _ { n + 1 } = S _ { n } + a _ { n + 1 }$ for natural numbers $n$,
$$n a _ { n + 1 } = 2 S _ { n } + ( n + 1 ) ^ { 3 } \quad \cdots (\text{ㄱ})$$
For natural numbers $n \geq 2$,
$$( n - 1 ) a _ { n } = 2 S _ { n - 1 } + n ^ { 3 } \quad \cdots (\text{ㄴ})$$
By subtracting (ㄴ) from (ㄱ), we obtain
$$n a _ { n + 1 } = ( n + 1 ) a _ { n } + \quad \text { (가) }$$
Dividing both sides by $n ( n + 1 )$,
$$\frac { a _ { n + 1 } } { n + 1 } = \frac { a _ { n } } { n } + \frac { ( \text{가} ) } { n ( n + 1 ) }$$
If $b _ { n } = \frac { a _ { n } } { n }$, then
$$b _ { n + 1 } = b _ { n } + \text{ (나) } \quad ( n \geq 2 )$$
so
$$b _ { n } = b _ { 2 } + \text{ (다) } \quad ( n \geq 3 )$$
When the expressions that go in (가), (나), and (다) are $f ( n ) , g ( n ) , h ( n )$ respectively, what is the value of $\frac { f ( 3 ) } { g ( 3 ) h ( 6 ) }$? [4 points]
(1) 30
(2) 36
(3) 42
(4) 48
(5) 54

What is the value of $\lim _ { n \rightarrow \infty } \frac { 5n ^ { 2 } + 1 } { 3n ^ { 2 } - 1 }$? [2 points]
(1) $\frac { 1 } { 3 }$
(2) $\frac { 2 } { 3 }$
(3) 1
(4) $\frac { 4 } { 3 }$
(5) $\frac { 5 } { 3 }$

The sequence $\left\{ a_n \right\}$ satisfies $a_1 = 4$ and $$a_{n+1} = n \cdot 2^n + \sum_{k=1}^{n} \frac{a_k}{k} \quad (n \geq 1)$$ The following is the process of finding the general term $a_n$.
From the given equation, $$a_n = (n-1) \cdot 2^{n-1} + \sum_{k=1}^{n-1} \frac{a_k}{k} \quad (n \geq 2)$$ Therefore, for natural numbers $n \geq 2$, $$a_{n+1} - a_n = \text{(가)} + \frac{a_n}{n}$$ so $$a_{n+1} = \frac{(n+1)a_n}{n} + \text{(가)}$$ If $b_n = \frac{a_n}{n}$, then $$b_{n+1} = b_n + \frac{(\text{가})}{n+1} \quad (n \geq 2)$$ and since $b_2 = 3$, $$b_n = \text{(나)} \quad (n \geq 2)$$ Therefore, $$a_n = \begin{cases} 4 & (n = 1) \\ n \times (\text{나}) & (n \geq 2) \end{cases}$$ If the expressions for (가) and (나) are $f(n)$ and $g(n)$, respectively, what is the value of $f(4) + g(7)$? [4 points]
(1) 90
(2) 95
(3) 100
(4) 105
(5) 110

The sequence $\left\{ a _ { n } \right\}$ has $a _ { 1 } = 4$ and satisfies
$$a _ { n + 1 } = n \cdot 2 ^ { n } + \sum _ { k = 1 } ^ { n } \frac { a _ { k } } { k } \quad ( n \geq 1 )$$
The following is the process of finding the general term $a _ { n }$.
From the given equation,
$$a _ { n } = ( n - 1 ) \cdot 2 ^ { n - 1 } + \sum _ { k = 1 } ^ { n - 1 } \frac { a _ { k } } { k } \quad ( n \geq 2 )$$
Therefore, for natural numbers $n \geq 2$,
$$a _ { n + 1 } - a _ { n } = \text { (a) } + \frac { a _ { n } } { n }$$
so
$$a _ { n + 1 } = \frac { ( n + 1 ) a _ { n } } { n } + \text { (a) }$$
If $b _ { n } = \frac { a _ { n } } { n }$, then
$$b _ { n + 1 } = b _ { n } + \frac { ( \text { a } ) } { n + 1 } ( n \geq 2 )$$
and since $b _ { 2 } = 3$,
$$b _ { n } = \text { (b) } \quad ( n \geq 2 )$$
Therefore,
$$a _ { n } = \left\{ \begin{array} { c c } 4 & ( n = 1 ) \\ n \times ( \boxed { ( \text{b} ) } ) & ( n \geq 2 ) \end{array} \right.$$
Let $f ( n )$ and $g ( n )$ be the expressions that fit (a) and (b), respectively. What is the value of $f ( 4 ) + g ( 7 )$? [4 points]
(1) 90
(2) 95
(3) 100
(4) 105
(5) 110

For the sequence $\left\{ a_n \right\}$, $$\sum_{n=1}^{\infty} \left( na_n - \frac{n^2 + 1}{2n + 1} \right) = 3$$ What is the value of $\lim_{n \rightarrow \infty} \left( a_n^2 + 2a_n + 2 \right)$? [4 points]
(1) $\frac{13}{4}$
(2) 3
(3) $\frac{11}{4}$
(4) $\frac{5}{2}$
(5) $\frac{9}{4}$

For a natural number $n$, the point $\mathrm { P } _ { n }$ on the coordinate plane is determined according to the following rules.
(a) The coordinates of the three points $\mathrm { P } _ { 1 } , \mathrm { P } _ { 2 } , \mathrm { P } _ { 3 }$ are $( - 1,0 ) , ( 1,0 )$, and $( - 1,2 )$, respectively.
(b) The midpoint of line segment $\mathrm { P } _ { n } \mathrm { P } _ { n + 1 }$ and the midpoint of line segment $\mathrm { P } _ { n + 2 } \mathrm { P } _ { n + 3 }$ are the same. For example, the coordinates of point $\mathrm { P } _ { 4 }$ are $( 1 , - 2 )$. When the coordinates of point $\mathrm { P } _ { 25 }$ are $( a , b )$, find the value of $a + b$. [4 points]

What is the value of $\lim _ { n \rightarrow \infty } \frac { 2 \times 3 ^ { n + 1 } + 5 } { 3 ^ { n } }$? [2 points]
(1) 6
(2) 7
(3) 8
(4) 9
(5) 10

A sequence $\left\{ a _ { n } \right\}$ with all positive terms has $a _ { 1 } = 10$ and satisfies $$\left( a _ { n + 1 } \right) ^ { n } = 10 \left( a _ { n } \right) ^ { n + 1 } \quad ( n \geq 1 )$$ The following is the process of finding the general term $a _ { n }$.
Taking the common logarithm of both sides of the given equation, $$n \log a _ { n + 1 } = ( n + 1 ) \log a _ { n } + 1$$ Dividing both sides by $n ( n + 1 )$, $$\frac { \log a _ { n + 1 } } { n + 1 } = \frac { \log a _ { n } } { n } + ( \text { (가) } )$$ Let $b _ { n } = \frac { \log a _ { n } } { n }$. Then $b _ { 1 } = 1$ and $$b _ { n + 1 } = b _ { n } + \text { (가) }$$ Finding the general term of the sequence $\left\{ b _ { n } \right\}$, $$b _ { n } = \text { (나) }$$ Therefore, $$\log a _ { n } = n \times \text { (나) }$$ Thus $a _ { n } = 10 ^ { n \times ( \text { (나) } ) }$.
When the expressions for (가) and (나) are $f ( n )$ and $g ( n )$ respectively, what is the value of $\frac { g ( 10 ) } { f ( 4 ) }$? [3 points]
(1) 38
(2) 40
(3) 42
(4) 44
(5) 46

A sequence $\left\{ a _ { n } \right\}$ with all positive terms has $a _ { 1 } = 10$ and satisfies
$$\left( a _ { n + 1 } \right) ^ { n } = 10 \left( a _ { n } \right) ^ { n + 1 } \quad ( n \geq 1 )$$
The following is the process of finding the general term $a _ { n }$.
Taking the common logarithm of both sides of the given equation:
$$n \log a _ { n + 1 } = ( n + 1 ) \log a _ { n } + 1$$
Dividing both sides by $n ( n + 1 )$:
$$\frac { \log a _ { n + 1 } } { n + 1 } = \frac { \log a _ { n } } { n } + ( \text{(가)} )$$
Let $b _ { n } = \frac { \log a _ { n } } { n }$. Then $b _ { 1 } = 1$ and
$$b _ { n + 1 } = b _ { n } + \text{(가)}$$
Finding the general term of the sequence $\left\{ b _ { n } \right\}$:
$$b _ { n } = \text{(나)}$$
Therefore,
$$\log a _ { n } = n \times \text{(나)}$$
Thus $a _ { n } = 10 ^ { n \times \text{(나)} }$.
Let $f ( n )$ and $g ( n )$ be the expressions that fit in (가) and (나), respectively. What is the value of $\frac { g ( 10 ) } { f ( 4 ) }$? [4 points]
(1) 38
(2) 40
(3) 42
(4) 44
(5) 46

What is the value of $\lim _ { n \rightarrow \infty } \frac { 4 n ^ { 2 } + 6 } { n ^ { 2 } + 3 n }$? [2 points]
(1) 1
(2) 2
(3) 3
(4) 4
(5) 5

The sequence $\left\{ a _ { n } \right\}$ satisfies $a _ { 1 } = 1$, and with $S _ { n } = \sum _ { k = 1 } ^ { n } a _ { k }$, $$a _ { n + 1 } = ( n + 1 ) S _ { n } + n ! \quad ( n \geq 1 )$$ The following is the process of finding the general term $a _ { n }$.
For a natural number $n$, since $a _ { n + 1 } = S _ { n + 1 } - S _ { n }$, by the given equation, $$S _ { n + 1 } = ( n + 2 ) S _ { n } + n ! \quad ( n \geq 1 )$$ Dividing both sides by $( n + 2 ) !$, $$\frac { S _ { n + 1 } } { ( n + 2 ) ! } = \frac { S _ { n } } { ( n + 1 ) ! } + \frac { 1 } { ( n + 1 ) ( n + 2 ) }$$ Let $b _ { n } = \frac { S _ { n } } { ( n + 1 ) ! }$. Then $b _ { 1 } = \frac { 1 } { 2 }$ and $$b _ { n + 1 } = b _ { n } + \frac { 1 } { ( n + 1 ) ( n + 2 ) }$$ Finding the general term of the sequence $\left\{ b _ { n } \right\}$, $$b _ { n } = \frac { ( \text{(가)} ) } { n + 1 }$$ Therefore, $$S _ { n } = \text{(가)} \times n!$$ Thus, $$a _ { n } = \text{(나)} \times ( n - 1 ) ! \quad ( n \geq 1 )$$ When the expressions that fit (가) and (나) are $f ( n )$ and $g ( n )$ respectively, what is the value of $f ( 7 ) + g ( 6 )$? [4 points]
(1) 44
(2) 41
(3) 38
(4) 35
(5) 32

For a natural number $n$, let $a _ { n }$ be the smallest natural number $m$ satisfying the following conditions. What is the value of $\sum _ { n = 1 } ^ { 10 } a _ { n }$? [4 points] (가) The coordinates of point A are $\left( 2 ^ { n } , 0 \right)$. (나) Let D be the point on the line passing through two points $\mathrm { B } ( 1,0 )$ and $\mathrm { C } \left( 2 ^ { m } , m \right)$ whose $x$-coordinate is $2 ^ { n }$. The area of triangle ABD is less than or equal to $\frac { m } { 2 }$.
(1) 109
(2) 111
(3) 113
(4) 115
(5) 117

For two sequences $\left\{ a _ { n } \right\}$ and $\left\{ b _ { n } \right\}$, $$\sum _ { n = 1 } ^ { \infty } a _ { n } = 4 , \quad \sum _ { n = 1 } ^ { \infty } b _ { n } = 10$$ find the value of $\sum _ { n = 1 } ^ { \infty } \left( a _ { n } + 5 b _ { n } \right)$. [3 points]

For a natural number $k$, $$a _ { k } = \lim _ { n \rightarrow \infty } \frac { \left( \frac { 6 } { k } \right) ^ { n + 1 } } { \left( \frac { 6 } { k } \right) ^ { n } + 1 }$$ Find the value of $\sum _ { k = 1 } ^ { 10 } k a _ { k }$. [4 points]

A sequence $\left\{ a _ { n } \right\}$ with all positive terms has $a _ { 1 } = a _ { 2 } = 1$, and with $S _ { n } = \sum _ { k = 1 } ^ { n } a _ { k }$, it satisfies $$a _ { n + 1 } = \frac { S _ { n } ^ { 2 } } { S _ { n - 1 } } + ( 2 n - 1 ) S _ { n } \quad ( n \geq 2 )$$ The following is the process of finding the general term $a _ { n }$. $$\begin{gathered} \text{Since } a _ { n + 1 } = S _ { n + 1 } - S _ { n } \text{, from the given equation we have} \\ S _ { n + 1 } = \frac { S _ { n } ^ { 2 } } { S _ { n - 1 } } + 2 n S _ { n } \quad ( n \geq 2 ) \end{gathered}$$ Dividing both sides by $S _ { n }$, we get $$\frac { S _ { n + 1 } } { S _ { n } } = \frac { S _ { n } } { S _ { n - 1 } } + 2 n$$ Let $b _ { n } = \frac { S _ { n + 1 } } { S _ { n } }$. Then $b _ { 1 } = 2$ and $$b _ { n } = b _ { n - 1 } + 2 n \quad ( n \geq 2 )$$ The general term of sequence $\left\{ b _ { n } \right\}$ is $$b _ { n } = \text { (가) } \times ( n + 1 ) \quad ( n \geq 1 )$$ Therefore, $$S _ { n } = ( \text{가} ) \times \{ ( n - 1 ) ! \} ^ { 2 } \quad ( n \geq 1 )$$ Thus $a _ { 1 } = 1$, and for $n \geq 2$, $$\begin{aligned} a _ { n } & = S _ { n } - S _ { n - 1 } \\ & = \text { (나) } \times \{ ( n - 2 ) ! \} ^ { 2 } \end{aligned}$$ Let $f ( n )$ and $g ( n )$ be the expressions that fit (가) and (나), respectively. What is the value of $f ( 10 ) + g ( 6 )$? [4 points]
(1) 110
(2) 125
(3) 140
(4) 155
(5) 170

A sequence $\left\{ a _ { n } \right\}$ with all positive terms has $a _ { 1 } = a _ { 2 } = 1$, and when $S _ { n } = \sum _ { k = 1 } ^ { n } a _ { k }$, $$a _ { n + 1 } = \frac { S _ { n } ^ { 2 } } { S _ { n - 1 } } + ( 2 n - 1 ) S _ { n } \quad ( n \geq 2 )$$ The following is the process of finding the general term $a _ { n }$.
Since $a _ { n + 1 } = S _ { n + 1 } - S _ { n }$, from the given equation we have $$S _ { n + 1 } = \frac { S _ { n } ^ { 2 } } { S _ { n - 1 } } + 2 n S _ { n } \quad ( n \geq 2 )$$ Dividing both sides by $S _ { n }$, $$\frac { S _ { n + 1 } } { S _ { n } } = \frac { S _ { n } } { S _ { n - 1 } } + 2 n$$ Let $b _ { n } = \frac { S _ { n + 1 } } { S _ { n } }$. Then $b _ { 1 } = 2$ and $$b _ { n } = b _ { n - 1 } + 2 n \quad ( n \geq 2 )$$ Finding the general term of the sequence $\left\{ b _ { n } \right\}$, $$b _ { n } = \text { (가) } \times ( n + 1 ) \quad ( n \geq 1 )$$ Therefore, $$S _ { n } = ( \text{가} ) \times \{ ( n - 1 ) ! \} ^ { 2 } \quad ( n \geq 1 )$$ Thus $a _ { 1 } = 1$, and for $n \geq 2$, $$\begin{aligned} a _ { n } & = S _ { n } - S _ { n - 1 } \\ & = \text { (나) } \times \{ ( n - 2 ) ! \} ^ { 2 } \end{aligned}$$ When the expressions for (가) and (나) are $f ( n )$ and $g ( n )$ respectively, what is the value of $f ( 10 ) + g ( 6 )$? [4 points]
(1) 110
(2) 125
(3) 140
(4) 155
(5) 170

For a natural number $n$, let $\mathrm { P } _ { n }$ be the point where the line $x = 4 ^ { n }$ meets the curve $y = \sqrt { x }$. Let $L _ { n }$ be the length of the segment $\mathrm { P } _ { n } \mathrm { P } _ { n + 1 }$. Find the value of $\lim _ { n \rightarrow \infty } \left( \frac { L _ { n + 1 } } { L _ { n } } \right) ^ { 2 }$. [4 points]