In the expansion of $( a + x ) ( 1 + x ) ^ { 4 }$, the sum of coefficients of odd-power terms of $x$ is 32. Then $a = $ $\_\_\_\_$ .

The coefficient of $x ^ { 4 }$ in the expansion of $\left( x ^ { 2 } + \frac { 2 } { x } \right) ^ { 3 }$ is
A. 10
B. 20
C. 40
D. 80

The coefficient of $x ^ { 3 }$ in the expansion of $\left( 1 + 2 x ^ { 2 } \right) ( 1 + x ) ^ { 4 }$ is
A. 12
B. 16
C. 20
D. 24

4. The coefficient of $x ^ { 3 }$ in the expansion of $\left( 1 + 2 x ^ { 2 } \right) ( 1 + x ) ^ { 4 }$ is
A. 12
B. 16
C. 20
D. 24

The coefficient of $x ^ { 3 } y ^ { 3 }$ in the expansion of $\left( x + \frac { y ^ { 2 } } { x } \right) ( x + y ) ^ { 5 }$ is
A. 5
B. 10
C. 15
D. 20

The constant term in the expansion of $\left( x ^ { 2 } + \frac { 2 } { x } \right) ^ { 6 }$ is $\_\_\_\_$ (answer with a number).

13. The coefficient of $x ^ { 2 } y ^ { 6 }$ in the expansion of $\left( 1 - \frac { y } { x } \right) ( x + y ) ^ { 8 }$ is $\_\_\_\_$ (answer with a number).

The coefficient of $x ^ { 3 }$ in the binomial expansion of $( x - \sqrt { x } ) ^ { 4 }$ is \_\_\_\_

For $n \in \mathbb{N}$ and $k \in \llbracket 0; n \rrbracket$, we denote by $\binom{n}{k}$ the binomial coefficient $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.
Let $p \in \mathbb{N}^*$. Show that $\binom{2p}{p}$ is an even integer.
Deduce that, if $n \in \mathbb{N}^*$ and $p \in \llbracket 1; n \rrbracket$, then $\binom{n+p}{p}\binom{n}{p}$ is an even integer.

For $n \in \mathbb{N}$ and $k \in \llbracket 0; n \rrbracket$, we denote by $\binom{n}{k}$ the binomial coefficient $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. The family $\left(K_n\right)_{n \in \mathbb{N}}$ is the orthonormal family defined in question II.E.
For all $n \in \mathbb{N}$, show that we can write: $$K_n = \sqrt{2n+1} \, \Lambda_n$$ where $\Lambda_n$ is a polynomial with integer coefficients that we will make explicit.
Among the coefficients of $\Lambda_n$, which ones are even?

Show that $\sum _ { k = 0 } ^ { n } \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k } = 1$.

Show that $\sum _ { k = 0 } ^ { n } k \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k } = n x$.

Show that $\sum _ { k = 0 } ^ { n } k ( k - 1 ) \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k } = n ( n - 1 ) x ^ { 2 }$.

Deduce from the previous questions that $$\sum _ { k = 0 } ^ { n } \left( x - \frac { k } { n } \right) ^ { 2 } \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k } = \frac { x ( 1 - x ) } { n } .$$

Let $n \in \mathbb { N } ^ { * }$ and $x \in [ 0,1 ]$. We consider the sum $$S ( x ) = \sum _ { k = 0 } ^ { n } \left| x - \frac { k } { n } \right| \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k } .$$ We denote

• $V$ the set of integers $k \in \{ 0 , \ldots , n \}$ such that $\left| x - \frac { k } { n } \right| \leqslant \frac { 1 } { \sqrt { n } }$,
• $W$ the set of integers $k \in \{ 0 , \ldots , n \}$ such that $\left| x - \frac { k } { n } \right| > \frac { 1 } { \sqrt { n } }$, and we set

$$S _ { V } ( x ) = \sum _ { k \in V } \left| x - \frac { k } { n } \right| \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k } \quad \text { and } \quad S _ { W } ( x ) = \sum _ { k \in W } \left| x - \frac { k } { n } \right| \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k } .$$ a) Show that $S _ { V } ( x ) \leqslant \frac { 1 } { \sqrt { n } }$. b) Show that $S _ { W } ( x ) \leqslant \frac { x ( 1 - x ) } { \sqrt { n } }$. c) Deduce that $S ( x ) \leqslant \frac { 5 } { 4 \sqrt { n } }$.

Let $n \in \mathbb { N } ^ { * }$ and $x \in [ 0,1 ]$. We consider the sum $$S ( x ) = \sum _ { k = 0 } ^ { n } \left| x - \frac { k } { n } \right| \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k } .$$ a) Write the Cauchy-Schwarz inequality in the space $\mathbb { R } ^ { n + 1 }$ equipped with its canonical inner product. b) Using question I.A.4, deduce that $S ( x ) \leqslant \frac { 1 } { 2 \sqrt { n } }$.

We are given a real sequence $\left(u_k\right)_{k \in \mathbb{N}}$ and we define for every integer $k \in \mathbb{N}$ $$v_k = \sum_{j=0}^{k} \binom{k}{j} u_j$$ Determine a matrix $Q \in \mathcal{M}_{n+1}(\mathbb{R})$ such that $$\left(\begin{array}{c} v_0 \\ v_1 \\ \vdots \\ v_n \end{array}\right) = Q \left(\begin{array}{c} u_0 \\ u_1 \\ \vdots \\ u_n \end{array}\right)$$

Deduce the inversion formula: for every integer $k \in \mathbb{N}$, $$u_k = \sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} v_j$$

We consider a real $\lambda$ and the sequence $\left(u_k = \lambda^k\right)_{k \in \mathbb{N}}$. What is the sequence $\left(v_k\right)_{k \in \mathbb{N}}$ defined by formula $$v_k = \sum_{j=0}^{k} \binom{k}{j} u_j \quad \text{(I.1)}$$ Then verify formula $$u_k = \sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} v_j \quad \text{(I.2)}$$

For an application $f : \mathbb{R}_+^* \rightarrow \mathbb{R}$ of class $\mathcal{C}^\infty$, we define the application $$\delta(f) : \left\{ \begin{array}{l} \mathbb{R}_+^* \rightarrow \mathbb{R} \\ x \mapsto f(x+1) - f(x) \end{array} \right.$$
For $n \in \mathbb{N}$ and $x > 0$, express $\left(\delta^n(f)\right)(x)$ using the binomial coefficients $\binom{n}{j}$ and the $f(x+j)$ (where the index $j$ belongs to $\llbracket 0, n \rrbracket$).

We introduce a uniformly distributed random variable $Z : \Omega \rightarrow \{-1,1\}^{n}$.
(a) Show that for $m \in \{0, \ldots, n-1\}$, we have $$\sum_{k=0}^{m} (n - 2k) \binom{n}{k} = n \binom{n-1}{m}.$$
(b) Deduce that for all $A \in \mathcal{M}_{n}(\{-1,1\})$, $$\mathbb{E}\left[g_{A}(Z)\right] = \frac{n^{2}}{2^{n-1}} \binom{n-1}{\left\lfloor \frac{n}{2} \right\rfloor}$$ where $\left\lfloor \frac{n}{2} \right\rfloor$ denotes the floor of $\frac{n}{2}$.

1. We denote $E_n = \operatorname{Card} \operatorname{MD}(n)$ and $\mathscr{I}_n$ the set of odd numbers in $\Delta_n$. a. Prove that for $n \geq 1$: $$E_{n+1} = \sum_{i \in \mathscr{I}_{n+1}} \binom{n}{i-1} E_{i-1} E_{n+1-i}$$ b. Deduce that for $n \geq 1$: $$2E_{n+1} = \sum_{i=0}^{n} \binom{n}{i} E_i E_{n-i}$$

Let $n \in \mathbb{N}$. Using the factorization $$( X + 1 ) ^ { 2 n } = ( X + 1 ) ^ { n } ( X + 1 ) ^ { n }$$ show that $$\sum _ { k = 0 } ^ { n } \binom { n } { k } ^ { 2 } = \binom { 2 n } { n }$$

Deduce $\forall n \in \mathbb { N } , C _ { n } = \frac { 1 } { n + 1 } \binom { 2 n } { n }$.