Exercise 4 — Candidates who have NOT followed the specialization course
We consider the sequence $(u_n)$ defined on $\mathbb{N}$ by: $$u_0 = 2 \text{ and for all natural integer } n,\quad u_{n+1} = \frac{u_n + 2}{2u_n + 1}.$$ We admit that for all natural integer $n$, $u_n > 0$.

1. a. Calculate $u_1, u_2, u_3, u_4$. An approximate value to $10^{-2}$ may be given. b. Verify that if $n$ is one of the integers $0,1,2,3,4$ then $u_n - 1$ has the same sign as $(-1)^n$. c. Establish that for all natural integer $n$, $u_{n+1} - 1 = \dfrac{-u_n + 1}{2u_n + 1}$. d. Prove by induction that for all natural integer $n$, $u_n - 1$ has the same sign as $(-1)^n$.
2. For all natural integer $n$, we set $v_n = \dfrac{u_n - 1}{u_n + 1}$. a. Establish that for all natural integer $n$, $v_{n+1} = \dfrac{-u_n + 1}{3u_n + 3}$. b. Prove that the sequence $(v_n)$ is a geometric sequence with ratio $-\dfrac{1}{3}$. Deduce the expression of $v_n$ as a function of $n$. c. We admit that for all natural integer $n$, $u_n = \dfrac{1 + v_n}{1 - v_n}$. Express $u_n$ as a function of $n$ and determine the limit of the sequence $(u_n)$.

Exercise 4 (Candidates who have not followed the mathematics specialization course)
Consider the sequence $( u _ { n } )$ defined by $u _ { 0 } = \frac { 1 } { 2 }$ and such that for every natural integer $n$,
$$u _ { n + 1 } = \frac { 3 u _ { n } } { 1 + 2 u _ { n } }$$

1. a. Calculate $u _ { 1 }$ and $u _ { 2 }$. b. Prove, by induction, that for every natural integer $n , 0 < u _ { n }$.
2. We admit that for every natural integer $n , u _ { n } < 1$. a. Prove that the sequence $\left( u _ { n } \right)$ is increasing. b. Prove that the sequence $( u _ { n } )$ converges.
3. Let $\left( v _ { n } \right)$ be the sequence defined, for every natural integer $n$, by $v _ { n } = \frac { u _ { n } } { 1 - u _ { n } }$. a. Show that the sequence $\left( v _ { n } \right)$ is a geometric sequence with ratio 3. b. Express for every natural integer $n , v _ { n }$ as a function of $n$. c. Deduce that, for every natural integer $n , u _ { n } = \frac { 3 ^ { n } } { 3 ^ { n } + 1 }$. d. Determine the limit of the sequence $\left( u _ { n } \right)$.

Exercise 2 (5 points) -- Common to all candidates

Part A

Let $(u_n)$ be the sequence defined by its first term $u_0$ and, for every natural number $n$, by the relation $$u_{n+1} = a u_n + b \quad (a \text{ and } b \text{ non-zero real numbers such that } a \neq 1).$$ We set, for every natural number $n$, $\quad v_n = u_n - \dfrac{b}{1-a}$.

1. Prove that the sequence $(v_n)$ is geometric with common ratio $a$.
2. Deduce that if $a$ belongs to the interval $]-1\,;\,1[$, then the sequence $(u_n)$ has limit $\dfrac{b}{1-a}$.

Part B

In March 2015, Max buys a green plant measuring 80 cm. He is advised to prune it every year, in March, by cutting a quarter of its height. The plant will then grow 30 cm over the following twelve months. As soon as he gets home, Max prunes his plant.

1. What will be the height of the plant in March 2016 before Max prunes it?
2. For every natural number $n$, we denote by $h_n$ the height of the plant, before pruning, in March of the year $(2015 + n)$. a. Justify that, for every natural number $n$, $\quad h_{n+1} = 0.75\,h_n + 30$. b. Conjecture using a calculator the direction of variation of the sequence $(h_n)$. Prove this conjecture (you may use a proof by induction). c. Is the sequence $(h_n)$ convergent? Justify your answer.

Exercise 2

Let $u$ be the sequence defined by $u _ { 0 } = 2$ and, for every natural integer $n$, by $$u _ { n + 1 } = 2 u _ { n } + 2 n ^ { 2 } - n .$$ We also consider the sequence $v$ defined, for every natural integer $n$, by $$v _ { n } = u _ { n } + 2 n ^ { 2 } + 3 n + 5$$

1. Here is an extract from a spreadsheet:
   

   	A	B	C
   1	$n$	$u$	$v$
   2	0	2	7
   3	1	4	14
   4	2	9	28
   5	3	24	56
   6	4	63
   7

   
   What formulas were written in cells C2 and B3 and copied downward to display the terms of the sequences $u$ and $v$?
2. Determine, by justifying, an expression of $v _ { n }$ and of $u _ { n }$ as a function of $n$ only.

Main topics covered: Numerical sequences; proof by induction.
We consider the sequences $(u_n)$ and $(v_n)$ defined by:
$$u_0 = 16 \quad; \quad v_0 = 5$$
and for any natural number $n$:
$$\left\{\begin{aligned} u_{n+1} & = \frac{3u_n + 2v_n}{5} \\ v_{n+1} & = \frac{u_n + v_n}{2} \end{aligned}\right.$$

1. Calculate $u_1$ and $v_1$.
2. We consider the sequence $(w_n)$ defined for any natural number $n$ by: $w_n = u_n - v_n$. a. Prove that the sequence $(w_n)$ is geometric with ratio 0.1.
   Deduce from this, for any natural number $n$, the expression of $w_n$ as a function of $n$. b. Specify the sign of the sequence $(w_n)$ and the limit of this sequence.
3. a. Prove that, for any natural number $n$, we have: $u_{n+1} - u_n = -0.4w_n$. b. Deduce that the sequence $(u_n)$ is decreasing.
   It can be shown in the same way that the sequence $(v_n)$ is increasing. We admit this result, and we note that we then have: for any natural number $n$, $v_n \geq v_0 = 5$. c. Prove by induction that, for any natural number $n$, we have: $u_n \geq 5$.
   Deduce that the sequence $(u_n)$ is convergent. We call $\ell$ the limit of $(u_n)$. It can be shown in the same way that the sequence $(v_n)$ is convergent. We admit this result, and we call $\ell'$ the limit of $(v_n)$.
4. a. Prove that $\ell = \ell'$. b. We consider the sequence $(c_n)$ defined for any natural number $n$ by: $c_n = 5u_n + 4v_n$. Prove that the sequence $(c_n)$ is constant, that is, for any natural number $n$, we have: $c_{n+1} = c_n$. Deduce that, for any natural number $n$, $c_n = 100$. c. Determine the common value of the limits $\ell$ and $\ell'$.

Part A

Let $\left(u_{n}\right)$ be the sequence defined by $u_{0} = 30$ and, for every natural integer $n$, $u_{n+1} = \frac{1}{2} u_{n} + 10$. Let $(v_{n})$ be the sequence defined for every natural integer $n$ by $v_{n} = u_{n} - 20$.

1. Calculate the exact values of $u_{1}$ and $u_{2}$.
2. Prove that the sequence $(v_{n})$ is geometric with common ratio $\frac{1}{2}$.
3. Express $v_{n}$ as a function of $n$ for every natural integer $n$.
4. Deduce that, for every natural integer $n$, $u_{n} = 20 + 10\left(\frac{1}{2}\right)^{n}$.
5. Determine the limit of the sequence $\left(u_{n}\right)$. Justify the answer.

Part B

Let $(w_{n})$ be the sequence defined for every natural integer $n$ by: $$\left\{\begin{array}{l} w_{0} = 45 \\ w_{n+1} = \frac{1}{2} w_{n} + \frac{1}{2} u_{n} + 7 \end{array}\right.$$

1. Show that $w_{1} = 44.5$.

We wish to write a function \texttt{suite}, in Python language, which returns the value of the term $w_{n}$ for a given value of $n$. We give below a proposal for this function \texttt{suite}. \begin{verbatim} def suite(n) : U=30 W=45 for i in range (1,n+1) : U=U/2+10 W=W/2+U/2+7 return W \end{verbatim}

1. The execution of \texttt{suite(1)} does not return the term $w_{1}$. How should the function \texttt{suite} be modified so that the execution of \texttt{suite(n)} returns the value of the term $w_{n}$?
2. a. Show, by induction on $n$, that for every natural integer $n$ we have: $$w_{n} = 10n\left(\frac{1}{2}\right)^{n} + 11\left(\frac{1}{2}\right)^{n} + 34$$ b. We admit that for every natural integer $n \geqslant 4$, we have: $0 \leqslant 10n\left(\frac{1}{2}\right)^{n} \leqslant \frac{10}{n}$. What can we deduce about the convergence of the sequence $\left(w_{n}\right)$?

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) $\frac { 1 } { 98 }$

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

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

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

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

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

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

Consider a sequence of positive numbers $a _ { 1 } , a _ { 2 } , a _ { 3 } , \cdots$ which satisfies
$$\begin{aligned} a _ { 1 } & = 1 , \quad a _ { 2 } = 10 , \\ \left( a _ { n } \right) ^ { 2 } a _ { n - 2 } & = \left( a _ { n - 1 } \right) ^ { 3 } \quad ( n = 3,4 , \cdots ) . \end{aligned}$$
We are to find $\lim _ { n \rightarrow \infty } a _ { n }$.
By finding the common logarithm of both sides of (1), we obtain
$$\mathbf { A } \log _ { 10 } a _ { n } + \log _ { 10 } a _ { n - 2 } = \mathbf { B } \log _ { 10 } a _ { n - 1 } .$$
When we set $b _ { n } = \log _ { 10 } a _ { n }$ $(n = 1,2 , \cdots)$, this equality is expressed as
$$\mathbf { A } b _ { n } + b _ { n - 2 } = \mathbf { B } b _ { n - 1 } .$$
By transforming (2), we have
$$b _ { n } - b _ { n - 1 } = \frac { 1 } { \mathbf { C } } \left( b _ { n - 1 } - b _ { n - 2 } \right) \quad ( n = 3,4 , \cdots ) ,$$
which gives
$$b _ { n } - b _ { n - 1 } = \left( \frac { 1 } { \mathbf { C } } \right) ^ { n - \mathbf { D } } \left( b _ { 2 } - b _ { 1 } \right) \quad ( n = 2,3 , \cdots ) .$$
Since $b _ { 1 } = \mathbf { E }$ and $b _ { 2 } = \mathbf { F }$, from (3) we get
$$b _ { n } = \sum _ { k = 2 } ^ { n } \left( \frac { 1 } { \mathbf { C } } \right) ^ { k - \mathbf { G } }$$
and hence
$$b _ { n } = \mathbf { H } - \left( \frac { 1 } { \mathbf { C } } \right) ^ { n - \mathbf{I} }$$
Finally, we obtain
$$\lim _ { n \rightarrow \infty } a _ { n } = \mathbf { J K L } .$$

Q1 We are to find the general term $a _ { n }$ of the sequence $\left\{ a _ { n } \right\}$ which is determined by the recurrence formula
$$a _ { 1 } = 18 , \quad a _ { n + 1 } - 12 a _ { n } + 3 ^ { n + 2 } = 0 \quad ( n = 1,2,3 , \cdots ) .$$
When we define a sequence $\left\{ b _ { n } \right\}$ by
$$b _ { n } = \frac { a _ { n } } { \mathbf{A}^n } \quad ( n = 1,2,3 , \cdots ) ,$$
$\left\{ b _ { n } \right\}$ satisfies
$$b _ { 1 } = \mathbf { B } , \quad b _ { n + 1 } - \mathbf { C } \; b _ { n } + \mathbf { D } = 0 \quad ( n = 1,2,3 , \cdots ) .$$
This recurrence formula can be transformed into
$$b _ { n + 1 } - \mathbf { E } = \mathbf{F} ( b_n - \mathbf{E} )$$
Next, when we define a sequence $\left\{ c _ { n } \right\}$ by
$$c _ { n } = b _ { n } - \mathbf { E } \quad ( n = 1,2,3 , \cdots ) ,$$
$\left\{ c _ { n } \right\}$ is a geometric progression such that the first term is $\mathbf{G}$ and the common ratio is $\mathbf{H}$.
Hence we have
$$a _ { n } = \mathbf { I } ^ { n } \left( \mathbf { J } \cdot \mathbf { K } ^ { n - 1 } + \mathbf { L } \right) \quad ( n = 1,2,3 , \cdots ) .$$

A sequence $< a _ { n } >$ satisfies $3 a _ { n + 1 } = a _ { n } + n$ (for all positive integers $n$) and $a _ { 1 } = 2$. Let the sequence $< b _ { n } >$ satisfy $b _ { n } = a _ { n } - \frac { n } { 2 } + \frac { 3 } { 4 }$. Select the correct options.
(1) $a _ { 2 } = 2$
(2) $b _ { 2 } = \frac { 3 } { 4 }$
(3) The sequence $< b _ { n } >$ is a geometric sequence with common ratio $\frac { 2 } { 3 }$
(4) For any positive integer $n$, $3 ^ { n } a _ { n }$ is always a positive integer
(5) $b _ { 10 } < 10 ^ { - 4 }$