Let $k \in \mathbb{N}^*$ and $(a_1, \ldots, a_k) \in (\mathbb{N}^*)^k$ a $k$-tuple of strictly positive integers. When $k \geq 2$, we assume they are coprime as a set. We define a function $P : \mathbb{N} \rightarrow \mathbb{C}$ by setting for all $n \in \mathbb{N}$: $$P(n) = \operatorname{Card}\left\{(n_1, \ldots, n_k) \in \mathbb{N}^k : n_1 a_1 + \cdots + n_k a_k = n\right\},$$ then we define the power series $F(x) = \sum_{n=0}^{\infty} P(n) x^n$.
Prove the equality $F(x) = \prod_{i=1}^{k} \frac{1}{1 - x^{a_i}}$ for all $x \in ]-1, 1[$.

Let $k \in \mathbb{N}^*$ and $(a_1, \ldots, a_k) \in (\mathbb{N}^*)^k$ a $k$-tuple of strictly positive integers. When $k \geq 2$, we assume they are coprime as a set. We define a function $P : \mathbb{N} \rightarrow \mathbb{C}$ by setting for all $n \in \mathbb{N}$: $$P(n) = \operatorname{Card}\left\{(n_1, \ldots, n_k) \in \mathbb{N}^k : n_1 a_1 + \cdots + n_k a_k = n\right\},$$ then we define the power series $F(x) = \sum_{n=0}^{\infty} P(n) x^n$.
Deduce that $P$ is a quasi-polynomial function.

Let $k \in \mathbb{N}^*$ and $(a_1, \ldots, a_k) \in (\mathbb{N}^*)^k$ a $k$-tuple of strictly positive integers. When $k \geq 2$, we assume they are coprime as a set. We define a function $P : \mathbb{N} \rightarrow \mathbb{C}$ by setting for all $n \in \mathbb{N}$: $$P(n) = \operatorname{Card}\left\{(n_1, \ldots, n_k) \in \mathbb{N}^k : n_1 a_1 + \cdots + n_k a_k = n\right\},$$ then we define the power series $F(x) = \sum_{n=0}^{\infty} P(n) x^n$.
Calculate the leading coefficient of $P$.

Let $k \in \mathbb{N}^*$ and $(a_1, \ldots, a_k) \in (\mathbb{N}^*)^k$ a $k$-tuple of strictly positive integers. When $k \geq 2$, we assume they are coprime as a set. We define a function $P : \mathbb{N} \rightarrow \mathbb{C}$ by setting for all $n \in \mathbb{N}$: $$P(n) = \operatorname{Card}\left\{(n_1, \ldots, n_k) \in \mathbb{N}^k : n_1 a_1 + \cdots + n_k a_k = n\right\}.$$ We assume $k = 2$. We assume in this question that $(a_1, a_2) = (2, 3)$. Construct a function $\phi : \mathbb{Z} \rightarrow \mathbb{Z}$ of period 6 such that $P(n) = \frac{n + \phi(n)}{6}$ for all $n \in \mathbb{N}$.

Let $k \in \mathbb{N}^*$ and $(a_1, \ldots, a_k) \in (\mathbb{N}^*)^k$ a $k$-tuple of strictly positive integers. When $k \geq 2$, we assume they are coprime as a set. We define a function $P : \mathbb{N} \rightarrow \mathbb{C}$ by setting for all $n \in \mathbb{N}$: $$P(n) = \operatorname{Card}\left\{(n_1, \ldots, n_k) \in \mathbb{N}^k : n_1 a_1 + \cdots + n_k a_k = n\right\}.$$ We assume $k = 2$. We set $a = a_1, b = a_2, \omega_a = \exp(2\mathrm{i}\pi/a), \omega_b = \exp(2\mathrm{i}\pi/b)$. From a partial fraction decomposition of the fraction $\frac{1}{(1 - x^a)(1 - x^b)}$, show the formula $$P(n) = \frac{1}{2a} + \frac{1}{2b} + \frac{n}{ab} + \frac{1}{a} \sum_{j=1}^{a-1} \frac{\omega_a^{-jn}}{1 - \omega_a^{jb}} + \frac{1}{b} \sum_{k=1}^{b-1} \frac{\omega_b^{-kn}}{1 - \omega_b^{ka}}$$ for all integer $n \geq 0$.

Let $k \in \mathbb{N}^*$ and $(a_1, \ldots, a_k) \in (\mathbb{N}^*)^k$ a $k$-tuple of strictly positive integers. When $k \geq 2$, we assume they are coprime as a set. We define a function $P : \mathbb{N} \rightarrow \mathbb{C}$ by setting for all $n \in \mathbb{N}$: $$P(n) = \operatorname{Card}\left\{(n_1, \ldots, n_k) \in \mathbb{N}^k : n_1 a_1 + \cdots + n_k a_k = n\right\}.$$ We assume $k = 2$. Prove that $$P(n) = \frac{n}{ab} - \left\{\frac{b^* n}{a}\right\} - \left\{\frac{a^* n}{b}\right\} + 1$$ for all integer $n \geq 0$, where $a^*$ and $b^*$ are integers satisfying $a^* a = 1$ modulo $b$ and $b^* b = 1$ modulo $a$ respectively.
Hint. One may use the formula $(*)$ for $b = 1$.

Let $n \in \mathbb { N } ^ { * }$. Justify that $\zeta ( 2 ) = \sum _ { k = 1 } ^ { + \infty } \frac { 1 } { k ^ { 2 } }$ is well-defined, then show that we can write
$$\sum _ { k = 1 } ^ { n } \frac { 1 } { k ^ { 2 } } = \frac { p _ { n } } { q _ { n } }$$
with $p _ { n } \in \mathbb { N } ^ { * }$ and $q _ { n } = d _ { n } ^ { 2 }$.

113. We classify the natural numbers in groups such that each group contains consecutive numbers, i.e., $\ldots, \{4,5,6\}, \{2,3\}, \{1\}, \ldots$ What is the sum of the numbers in the group containing 350?
$$4125 \quad (1) \qquad 4050 \quad (2) \qquad 4015 \quad (3) \qquad 3980 \quad (4)$$

For any $n \geq 5$, the value of $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^n - 1}$ lies between
(A) 0 and $\frac{n}{2}$
(B) $\frac{n}{2}$ and $n$
(C) $n$ and $2n$
(D) none of the above

For any $n \geq 5$, the value of $1 + \frac { 1 } { 2 } + \frac { 1 } { 3 } + \cdots + \frac { 1 } { 2 ^ { n } - 1 }$ lies between
(A) 0 and $\frac { n } { 2 }$
(B) $\frac { n } { 2 }$ and $n$
(C) $n$ and $2 n$
(D) none of the above

What is the limit of $\sum _ { k = 1 } ^ { n } \frac { e ^ { - k / n } } { n }$ as $n$ tends to $\infty$ ?
(A) The limit does not exist.
(B) $\infty$
(C) $1 - e ^ { - 1 }$
(D) $e ^ { - 0.5 }$

The integral part of $\sum _ { n = 2 } ^ { 9999 } \frac { 1 } { \sqrt { n } }$ equals
(A) 196
(B) 197
(C) 198
(D) 199 .

Let $a _ { n }$ be the number of subsets of $\{ 1,2 , \ldots , n \}$ that do not contain any two consecutive numbers. Then
(A) $a _ { n } = a _ { n - 1 } + a _ { n - 2 }$
(B) $a _ { n } = 2 a _ { n - 1 }$
(C) $a _ { n } = a _ { n - 1 } - a _ { n - 2 }$
(D) $a _ { n } = a _ { n - 1 } + 2 a _ { n - 2 }$.

Consider the sequence $1,2,2,3,3,3,4,4,4,4,5,5,5,5,5 , \ldots$ obtained by writing one 1 , two 2's, three 3's and so on. What is the $2020 ^ { \text {th} }$ term in the sequence?
(A) 62
(B) 63
(C) 64
(D) 65

Define $S _ { n } = \frac { 1 } { 2 } \cdot \frac { 3 } { 4 } \cdots \cdot \frac { 2 n - 1 } { 2 n }$ where $n$ is a positive integer. Then
(A) $S _ { n } < \frac { 1 } { \sqrt { 4 n + 2 } }$ for some $n > 2$.
(B) $S _ { n } < \frac { 1 } { \sqrt { 2 n + 1 } }$ for all $n \geq 2$.
(C) $S _ { n } < \frac { 1 } { \sqrt { 2 n + 5 } }$ for all $n \geq 2$.
(D) $S _ { n } > \frac { 1 } { \sqrt { 4 n + 2 } }$ for all $n \geq 2$.

For any $n \geq 5$, the value of $1 + \frac { 1 } { 2 } + \frac { 1 } { 3 } + \cdots + \frac { 1 } { 2 ^ { n } - 1 }$ lies between
(a) 0 and $n / 2$.
(B) $n / 2$ and $n$.
(C) $n$ and $2 n$.
(D) none of the above.

Let $g ( x ) = \log f ( x )$ where $f ( x )$ is a twice differentiable positive function on $( 0 , \infty )$ such that $f ( x + 1 ) = x f ( x )$. Then, for $N = 1,2,3 , \ldots$,
$$g ^ { \prime \prime } \left( N + \frac { 1 } { 2 } \right) - g ^ { \prime \prime } \left( \frac { 1 } { 2 } \right) =$$
(A) $- 4 \left\{ 1 + \frac { 1 } { 9 } + \frac { 1 } { 25 } + \cdots + \frac { 1 } { ( 2 N - 1 ) ^ { 2 } } \right\}$
(B) $4 \left\{ 1 + \frac { 1 } { 9 } + \frac { 1 } { 25 } + \cdots + \frac { 1 } { ( 2 N - 1 ) ^ { 2 } } \right\}$
(C) $- 4 \left\{ 1 + \frac { 1 } { 9 } + \frac { 1 } { 25 } + \cdots + \frac { 1 } { ( 2 N + 1 ) ^ { 2 } } \right\}$
(D) $4 \left\{ 1 + \frac { 1 } { 9 } + \frac { 1 } { 25 } + \cdots + \frac { 1 } { ( 2 N + 1 ) ^ { 2 } } \right\}$

If the sum $\frac { 3 } { 1 ^ { 2 } } + \frac { 5 } { 1 ^ { 2 } + 2 ^ { 2 } } + \frac { 7 } { 1 ^ { 2 } + 2 ^ { 2 } + 3 ^ { 2 } } + \ldots + $ up to 20 terms is equal to $\frac { k } { 21 }$, then $k$ is equal to
(1) 240
(2) 120
(3) 60
(4) 180

The greatest positive integer $k$, for which $49 ^ { k } + 1$ is a factor of the sum $49 ^ { 125 } + 49 ^ { 124 } + \ldots + 49 ^ { 2 } + 49 + 1$, is
(1) 32
(2) 63
(3) 60
(4) 35

If $\cot ^ { - 1 } ( \alpha ) = \cot ^ { - 1 } 2 + \cot ^ { - 1 } 8 + \cot ^ { - 1 } 18 + \cot ^ { - 1 } 32 + \ldots$ upto 100 terms, then $\alpha$ is:
(1) 1.01
(2) 1.00
(3) 1.02
(4) 1.03

Let $\left\{ a _ { n } \right\} _ { n = 0 } ^ { \infty }$ be a sequence such that $a _ { 0 } = a _ { 1 } = 0$ and $a _ { n + 2 } = 2 a _ { n + 1 } - a _ { n } + 1$ for all $n \geq 0$. Then, $\sum _ { n = 2 } ^ { \infty } \frac { a _ { n } } { 7 ^ { n } }$ is equal to
(1) $\frac { 6 } { 343 }$
(2) $\frac { 7 } { 216 }$
(3) $\frac { 8 } { 343 }$
(4) $\frac { 49 } { 216 }$

$\sum_{r=1}^{20} (r^2 + 1) \cdot r!$ is equal to
(1) $22! - 21!$
(2) $22! - 2 \cdot 21!$
(3) $21! - 2 \cdot 20!$
(4) $21! - 20!$

The remainder on dividing $1 + 3 + 3 ^ { 2 } + 3 ^ { 3 } + \ldots + 3 ^ { 2021 }$ by 50 is $\_\_\_\_$.

If $\operatorname{gcd}(m, n) = 1$ and $1^{2} - 2^{2} + 3^{2} - 4^{2} + \ldots + (2021)^{2} - (2022)^{2} + (2023)^{2} = 1012m^{2}n$ then $m^{2} - n^{2}$ is equal to
(1) 240
(2) 200
(3) 220
(4) 180

Let $a _ { 1 } = b _ { 1 } = 1$ and $a _ { n } = a _ { n - 1 } + ( n - 1 ) , b _ { n } = b _ { n - 1 } + \mathrm { a } _ { n - 1 } , \forall n \geq 2$. If $\mathrm { S } = \sum _ { \mathrm { n } = 1 } ^ { 10 } \left( \frac { b _ { n } } { 2 ^ { n } } \right)$ and $\mathrm { T } = \sum _ { n = 1 } ^ { 8 } \frac { \mathrm { n } } { 2 ^ { n - 1 } }$ then $2 ^ { 7 } ( 2 S - T )$ is equal to $\_\_\_\_$