Given the sequence $\{a_n\}$ satisfies $a_{n+2} = qa_n$ (where q is a real number and $q \neq 1$), $n \in \mathbb{N}^*$, $a_1 = 1$, $a_2 = 2$, and $a_2 + a_3$, $a_3 + a_4$, $a_4 + a_5$ form an arithmetic sequence.
(I) Find the value of q and the general term formula of $\{a_n\}$;
(II) Let $b_n = \frac{\log_2 a_{2n}}{a_{2n-1}}$, $n \in \mathbb{N}^*$. Find the sum of the first n terms of the sequence $\{b_n\}$.

19. (This question is worth 13 points) Let $S _ { n }$ be the sum of the first $n$ terms of the sequence $\left\{ a _ { n } \right\}$. Given $a _ { 1 } = 1 , a _ { 2 } = 2$, and $a _ { n + 2 } = 3 S _ { n } - S _ { n + 1 } , n \in \mathbb { N } ^ { * }$. (I) Prove that: $a _ { n + 2 } = 3 a _ { n }$ (II) Find $\mathrm { S } _ { \mathrm { n } }$

20. (This question is worth 13 points) Given the sequence $\left\{ a _ { n } \right\}$ satisfying: $a _ { 1 } \in \mathbf { N } ^ { * } , a _ { 1 } \leqslant 36$, and $a _ { n + 1 } = \left\{ \begin{array} { l } 2 a _ { n } , a _ { n } \leqslant 18 , \\ 2 a _ { n } - 36 , a _ { n } > 18 \end{array} ( n = 1,2 , \ldots ) \right.$. Let the set $M = \left\{ a _ { n } \mid n \in \mathbf { N } ^ { * } \right\}$. (I) If $a _ { 1

20. (This question is worth 15 points) Given that the sequence $\left\{ a _ { n } \right\}$ satisfies $a _ { 1 } = \frac { 1 } { 2 }$ and $a _ { n + 1 } = a _ { n } - a _ { n } ^ { 2 }$ ( $n \in \mathbb{N

$\lim _ { n \rightarrow \infty } \frac { n + 1 } { 3 n - 1 } =$ $\_\_\_\_$

17.
(1)
By mathematical induction, we can deduce that
$$a _ { n } = \begin{cases} \frac { 3 n - 1 } { 2 } & \text{if } 2 \nmid n \\ \frac { 2 n - 2 } { 2 } & \text{if } 2 \mid n \end{cases}$$
Thus $b _ { n } = a _ { 2 n } = 3 n - 1$ for $n \in \mathbb { Z } ^ { + }$, with $b _ { 1 } = 2, b _ { 2 } = 5$.
(2)
$$\begin{gathered} \sum _ { k = 1 } ^ { 20 } a _ { k } = \sum _ { k = 1 } ^ { 10 } a _ { 2 k - 1 } + \sum _ { k = 1 } ^ { 10 } a _ { 2 k } \\ = \sum _ { k = 1 } ^ { 10 } ( 3 k - 2 ) + \sum _ { k = 1 } ^ { 10 } ( 3 k - 1 ) \\ = 6 \sum _ { k = 1 } ^ { 10 } k - 30 = 300 \end{gathered}$$

17. (10 points) Let $S _ { n }$ denote the sum of the first $n$ terms of the sequence $\left\{ a _ { n } \right\}$. Given that $a _ { 1 } = 1$ and $\left\{ \frac { S _ { n } } { a _ { n } } \right\}$ is an arithmetic sequence with common difference $\frac { 1 } { 3 }$.
(1) Find the general term formula for $\left\{ a _ { n } \right\}$;
(2) Prove that $\frac { 1 } { a _ { 1 } } + \frac { 1 } { a _ { 2 } } + \cdots + \frac { 1 } { a _ { n } } < 2$ .

In the sequence $\left\{ a_{n} \right\}$ , $a_{2} = 1$ . Let $S_{n}$ be the sum of the first $n$ terms of $\left\{ a_{n} \right\}$ . $2S_{n} = na_{n}$ .
(1) Find the general term formula for $\left\{ a_{n} \right\}$ ;
(2) Find the sum $T_{n}$ of the first $n$ terms of the sequence $\left\{ \frac{a_{n} + 1}{2^{n}} \right\}$ .

Let $k \in \mathbb { N } ^ { * }$. We denote by $c _ { k }$ the positive real number such that: $c _ { k } \int _ { 0 } ^ { 2 \pi } \left( \frac { 1 + \cos t } { 2 } \right) ^ { k } d t = 1$, and for every real $t$ we set $R _ { k } ( t ) = c _ { k } \left( \frac { 1 + \cos t } { 2 } \right) ^ { k }$.
Let $\epsilon \in ] 0 , \pi [$. We set: $d _ { k } ( \epsilon ) = \sup _ { t \in [ \epsilon , 2 \pi - \epsilon ] } R _ { k } ( t )$. Prove then that $$\lim _ { k \rightarrow + \infty } d _ { k } ( \epsilon ) = 0$$

Let $k \in \mathbb { N } ^ { * }$. We denote by $c _ { k }$ the positive real number such that: $c _ { k } \int _ { 0 } ^ { 2 \pi } \left( \frac { 1 + \cos t } { 2 } \right) ^ { k } d t = 1$, and for every real $t$ we set $R _ { k } ( t ) = c _ { k } \left( \frac { 1 + \cos t } { 2 } \right) ^ { k }$.
Let $\epsilon \in ] 0 , \pi [$, and $k \in \mathbb { N }$. Prove that for every $h \in C _ { 2 \pi } ( \mathbb { R } ; \mathbb { C } )$ that is of class $C ^ { 1 }$ on $\mathbb { R }$ and every real $u$, we have: $$\int _ { 0 } ^ { 2 \pi } R _ { k } ( u - t ) h ( t ) d t = \int _ { 0 } ^ { 2 \pi } R _ { k } \left( t _ { 1 } \right) h \left( u - t _ { 1 } \right) d t _ { 1 }$$ and $$\left| \int _ { 0 } ^ { 2 \pi } R _ { k } ( u - t ) h ( t ) d t - h ( u ) \right| \leq 2 \left\| h ^ { \prime } \right\| \epsilon + 4 \pi \| h \| d _ { k } ( \epsilon )$$ (We recall that $\int _ { 0 } ^ { 2 \pi } R _ { k } \left( t _ { 1 } \right) d t _ { 1 } = 1$ and that $\| h \|$ is defined at the beginning of the problem statement. To establish the inequality, one may use that $h \left( u - t _ { 1 } \right) = h \left( u - t _ { 1 } + 2 \pi \right)$ when $t _ { 1 } \in [ 2 \pi - \epsilon , 2 \pi ]$).

We consider the sequence $x = (x_n)_{n \geqslant 0}$ defined by $$x_0 = 1, \quad x_1 = 1, \quad x_2 = 1, \quad x_3 = 0, \quad \text{and} \quad \forall n \in \mathbb{N}, x_{n+4} = x_{n+3} - 2x_{n+1}$$
We decide to modify only the value of $x_0$, by setting this time $x_0 = \frac{1}{2}$.
With this modification, quickly redo the study of questions II.C.2 and II.C.3.

Let $\left( a _ { n } \right) _ { n \geqslant 0 }$ be a real sequence with $\sum a_n x^n$ having radius of convergence 1, with sum $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ satisfying $f(x) \sim \frac{1}{1-x}$ as $x \to 1^-$, and with $a_n \geqslant 0$ for all $n \in \mathbb{N}$. We denote $\widetilde{a}_n = \frac{A_n}{n+1}$ where $A_n = \sum_{k=0}^n a_k$.
Conclude (i.e., prove property II.3: $\lim_{n \to \infty} \widetilde{a}_n = 1$).

Let $n \in \mathbb { N }$.
Show that we can order the zeros of $\varphi _ { n }$, that is, there exists a strictly increasing sequence $\left( \alpha _ { k } ^ { ( n ) } \right) _ { k \in \mathbb { N } }$ of zeros of $\varphi _ { n }$ such that $\varphi _ { n }$ does not vanish on $] 0 , \alpha _ { 0 } ^ { ( n ) } [$ and on every interval $] \alpha _ { k } ^ { ( n ) } , \alpha _ { k + 1 } ^ { ( n ) } [$ with $k$ in $\mathbb { N }$ and that $\lim _ { k \rightarrow \infty } \alpha _ { k } ^ { ( n ) } = + \infty$.
Construct the sequence $\left( \alpha _ { k } ^ { ( n ) } \right) _ { k \in \mathbb { N } }$ by induction on $k$ by showing that the set $\mathcal { Z } _ { k }$ of zeros of $\varphi _ { n }$ in the interval $] \alpha _ { k } ^ { ( n ) } , + \infty [$ has a smallest element.

Let $\left( a _ { n } \right) _ { n \in \mathbb { N } }$ be a complex sequence. We assume that the series $\sum a _ { n }$ converges. For $n \in \mathbb { N }$, we denote $r _ { n } = \sum _ { k = n + 1 } ^ { + \infty } a _ { k }$ and we define the functions $s _ { n }$ and $s$ from $[ 0,1 ]$ to $\mathbb { C }$ by $s _ { n } ( x ) = \sum _ { k = 0 } ^ { n } a _ { k } x ^ { k }$ and $s ( x ) = \sum _ { k = 0 } ^ { + \infty } a _ { k } x ^ { k }$.
Justify the existence of $s$.

Let $\left( a _ { n } \right) _ { n \in \mathbb { N } }$ be a complex sequence. We assume that the series $\sum a _ { n }$ converges. For $n \in \mathbb { N }$, we denote $r _ { n } = \sum _ { k = n + 1 } ^ { + \infty } a _ { k }$ and we define the functions $s _ { n }$ and $s$ from $[ 0,1 ]$ to $\mathbb { C }$ by $s _ { n } ( x ) = \sum _ { k = 0 } ^ { n } a _ { k } x ^ { k }$ and $s ( x ) = \sum _ { k = 0 } ^ { + \infty } a _ { k } x ^ { k }$.
Let $x \in [ 0,1 ]$ and $n \in \mathbb { N } ^ { * }$. Show $$s ( x ) - s _ { n } ( x ) = r _ { n } x ^ { n + 1 } - \sum _ { k = n + 1 } ^ { + \infty } r _ { k } \left( x ^ { k } - x ^ { k + 1 } \right)$$

Let $\left( a _ { n } \right) _ { n \in \mathbb { N } }$ be a complex sequence. We assume that the series $\sum a _ { n }$ converges. For $n \in \mathbb { N }$, we denote $r _ { n } = \sum _ { k = n + 1 } ^ { + \infty } a _ { k }$ and we define the functions $s _ { n }$ and $s$ from $[ 0,1 ]$ to $\mathbb { C }$ by $s _ { n } ( x ) = \sum _ { k = 0 } ^ { n } a _ { k } x ^ { k }$ and $s ( x ) = \sum _ { k = 0 } ^ { + \infty } a _ { k } x ^ { k }$.
Show that $s$ is continuous on $[ 0,1 ]$.
For continuity at 1, fix $\varepsilon > 0$ and show that if the natural integer $N$ satisfies $\left| r _ { n } \right| \leqslant \varepsilon$ for all $n \geqslant N$, then $\left| s ( x ) - s _ { N } ( x ) \right| \leqslant 2 \varepsilon$ for all $x \in [ 0,1 ]$. Then bound the modulus of $s ( x ) - s ( 1 ) = \left( s ( x ) - s _ { N } ( x ) \right) + \left( s _ { N } ( x ) - s _ { N } ( 1 ) \right) + \left( s _ { N } ( 1 ) - s ( 1 ) \right)$.

Let $\left( a _ { n } \right) _ { n \in \mathbb { N } }$ be a complex sequence. We assume that the series $\sum a _ { n }$ converges. For $n \in \mathbb { N }$, we denote $r _ { n } = \sum _ { k = n + 1 } ^ { + \infty } a _ { k }$ and we define the functions $s _ { n }$ and $s$ from $[ 0,1 ]$ to $\mathbb { C }$ by $s _ { n } ( x ) = \sum _ { k = 0 } ^ { n } a _ { k } x ^ { k }$ and $s ( x ) = \sum _ { k = 0 } ^ { + \infty } a _ { k } x ^ { k }$.
Application: recover the result from question II.B.3.

Let $\theta \in \mathbb { R }$. Determine the power series expansion of the function $$x \mapsto \frac { 1 - x ^ { 2 } } { x ^ { 2 } - 2 x \cos \theta + 1 }$$ on an interval to be specified.

Let $f : \mathbb { R } \rightarrow \mathbb { R }$ be a $2 \pi$-periodic function of class $\mathcal { C } ^ { 1 }$. We consider the Fourier series of $f$ in cosines and sines, denoted $$c _ { 0 } + \sum _ { n \geq 1 } \left( a _ { n } \cos ( n t ) + b _ { n } \sin ( n t ) \right)$$
Show that, for all $x \in ] - 1,1 [$ and all $t \in \mathbb { R }$, $$c _ { 0 } + \sum _ { n = 1 } ^ { + \infty } \left( a _ { n } \cos ( n t ) + b _ { n } \sin ( n t ) \right) x ^ { n } = \frac { 1 } { 2 \pi } \int _ { 0 } ^ { 2 \pi } \frac { \left( 1 - x ^ { 2 } \right) f ( u ) } { x ^ { 2 } - 2 x \cos ( t - u ) + 1 } \mathrm { ~d} u$$

Let $f : \mathbb { R } \rightarrow \mathbb { R }$ be a $2 \pi$-periodic function of class $\mathcal { C } ^ { 1 }$. We consider the Fourier series of $f$ in cosines and sines, denoted $$c _ { 0 } + \sum _ { n \geq 1 } \left( a _ { n } \cos ( n t ) + b _ { n } \sin ( n t ) \right)$$
Deduce that, for all $t \in \mathbb { R }$, $$f ( t ) = \lim _ { x \rightarrow 1 ^ { - } } \frac { 1 } { 2 \pi } \int _ { 0 } ^ { 2 \pi } \frac { \left( 1 - x ^ { 2 } \right) f ( u ) } { x ^ { 2 } - 2 x \cos ( t - u ) + 1 } \mathrm { ~d} u$$

Let $a$ and $b$ be two complex numbers such that $(a,b) \neq (0,0)$. We say that a complex sequence $U = (u_n)_{n \in \mathbb{N}}$ satisfies the recurrence relation $(E_{a,b})$ if $$\forall n \in \mathbb{N}, \quad u_{n+2} = 2a u_{n+1} + b u_n$$ We assume that $a^2 + b \neq 0$. We denote $d = R(a^2 + b)$. We call $W$ the sequence $W = ((a+d)^n)_{n \in \mathbb{N}}$ and $W'$ the sequence $W' = ((a-d)^n)_{n \in \mathbb{N}}$. Show that $U$ satisfies $E_{a,b}$ if and only if $U \in \operatorname{Vect}(W, W')$. Determine $U$ satisfying $E_{a,b}$ and the initial conditions $u_0 = 0$ and $u_1 = 1$, as a function of $d$, $W$ and $W'$.

Let $a$ and $b$ be two complex numbers such that $(a,b) \neq (0,0)$. We say that a complex sequence $U = (u_n)_{n \in \mathbb{N}}$ satisfies the recurrence relation $(E_{a,b})$ if $$\forall n \in \mathbb{N}, \quad u_{n+2} = 2a u_{n+1} + b u_n$$ We assume that $a^2 + b = 0$ and $a \neq 0$. We denote $W$ and $W'$ the sequences $W = (a^n)_{n \in \mathbb{N}}$ and $W' = (na^n)_{n \in \mathbb{N}}$. Show that $U$ satisfies $E_{a,b}$ if and only if $U \in \operatorname{Vect}(W, W')$. Determine $U$ satisfying $E_{a,b}$ and the initial conditions $u_0 = 0$ and $u_1 = 1$, as a function of $a$, $W$ and $W'$.

We denote $V_n(z) = U_{n+1}(z, -1)$ for all $z \in \mathbb{C}$ and $n \in \mathbb{N}$, where $U(a,b) = (U_n(a,b))_{n \in \mathbb{N}}$ is the unique sequence satisfying $E_{a,b}$ with initial conditions $U_0(a,b) = 0$ and $U_1(a,b) = 1$. Explicitly write $V_1(z)$, $V_2(z)$ and $V_3(z)$ and determine their roots in $\mathbb{C}$.

We denote $V_n(z) = U_{n+1}(z, -1)$ for all $z \in \mathbb{C}$ and $n \in \mathbb{N}$, where $U(a,b) = (U_n(a,b))_{n \in \mathbb{N}}$ is the unique sequence satisfying $E_{a,b}$ with initial conditions $U_0(a,b) = 0$ and $U_1(a,b) = 1$. Show that, for all $z \in \mathbb{C}$ and $n \in \mathbb{N}$, we have $$V_n(z) = \sum_{j=0}^{\lfloor n/2 \rfloor} \binom{n-j}{j} (2z)^{n-2j} (-1)^j$$ One may proceed by induction.

We use the notation $R$ introduced in part I. Let $z \in \mathbb{C}$ such that $z^2 \neq 1$. We denote $$r = \left|R\left(z^2 - 1\right)\right|, \quad s = \left|z + R\left(z^2 - 1\right)\right|, \quad t = \left|z - R\left(z^2 - 1\right)\right|, \quad h = \max(s,t)$$ We also denote $V_n(z) = U_{n+1}(z,-1)$ for all $z \in \mathbb{C}$ and $n \in \mathbb{N}$. Prove that, for all $n \in \mathbb{N}$, $$\left|V_n(z)\right| \leqslant \frac{h^{n+1}}{r}$$