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

Let $n \in \mathbb{N}^*$ and $$T _ { n } ( X ) = \sum _ { p = 0 } ^ { \lfloor n / 2 \rfloor } ( - 1 ) ^ { p } \binom { n } { 2 p } X ^ { n - 2 p } \left( 1 - X ^ { 2 } \right) ^ { p }.$$ By expanding $( 1 + x ) ^ { n }$ for two appropriately chosen real numbers $x$, show that $$\sum _ { p = 0 } ^ { \lfloor n / 2 \rfloor } \binom { n } { 2 p } = 2 ^ { n - 1 }.$$

Show that, for all $n \in \mathbb { N } ^ { * }$,
$$\frac { 4 ^ { n } } { 2 n } \leqslant \binom { 2 n } { n } < 4 ^ { n } .$$

The number $$\left( \frac { 2 ^ { 10 } } { 11 } \right) ^ { 11 }$$ is
(A) strictly larger than $\binom{10}{1}^2 \binom{10}{2}^2 \binom{10}{3}^2 \binom{10}{4}^2 \binom{10}{5}$
(B) strictly larger than $\binom{10}{1}^2 \binom{10}{2}^2 \binom{10}{3}^2 \binom{10}{4}^2$ but strictly smaller than $\binom{10}{1}^2 \binom{10}{2}^2 \binom{10}{3}^2 \binom{10}{4}^2 \binom{10}{5}$
(C) less than or equal to $\binom{10}{1}^2 \binom{10}{2}^2 \binom{10}{3}^2 \binom{10}{4}^2$
(D) equal to $\binom{10}{1}^2 \binom{10}{2}^2 \binom{10}{3}^2 \binom{10}{4}^2 \binom{10}{5}$

The number $$\left( \frac { 2 ^ { 10 } } { 11 } \right) ^ { 11 }$$ is
(A) strictly larger than $\binom { 10 } { 1 } ^ { 2 } \binom { 10 } { 2 } ^ { 2 } \binom { 10 } { 3 } ^ { 2 } \binom { 10 } { 4 } ^ { 2 } \binom { 10 } { 5 }$
(B) strictly larger than $\binom { 10 } { 1 } ^ { 2 } \binom { 10 } { 2 } ^ { 2 } \binom { 10 } { 3 } ^ { 2 } \binom { 10 } { 4 } ^ { 2 }$ but strictly smaller than $\binom { 10 } { 1 } ^ { 2 } \binom { 10 } { 2 } ^ { 2 } \binom { 10 } { 3 } ^ { 2 } \binom { 10 } { 4 } ^ { 2 } \binom { 10 } { 5 }$
(C) less than or equal to $\binom { 10 } { 1 } ^ { 2 } \binom { 10 } { 2 } ^ { 2 } \binom { 10 } { 3 } ^ { 2 } \binom { 10 } { 4 } ^ { 2 }$
(D) equal to $\binom { 10 } { 1 } ^ { 2 } \binom { 10 } { 2 } ^ { 2 } \binom { 10 } { 3 } ^ { 2 } \binom { 10 } { 4 } ^ { 2 } \binom { 10 } { 5 }$

3. Let n be any positive integer. Prove that:
$$\sum _ { k = 0 } ^ { m } \frac { \binom { 2 n - k } { k } } { \binom { 2 n - k } { n } } \cdot \frac { ( 2 n - 4 k + 1 ) } { ( 2 n - 2 k + 1 ) } 2 ^ { n - 2 k } = \frac { \binom { n } { m } } { \binom { 2 n - 2 m } { n - m } } 2 ^ { n - 2 m }$$
for each nonnegative integer $\mathrm { m } \leq \mathrm { n }$. (Here $\left. \binom { p } { q } = \square ^ { p } C _ { q } \right)$

Prove that $$\begin{aligned} & 2 ^ { k } \binom { n } { 0 } \binom { n } { k } - 2 ^ { n - 1 } \binom { n } { 1 } \binom { n - 1 } { k - 1 } + 2 ^ { k - 2 } \\ & \binom { n } { 2 } \binom { n - 2 } { k - 2 } - \ldots \ldots \ldots ( - 1 ) ^ { k } \binom { n } { k } \binom { n - k } { 0 } = \binom { n } { k } . \end{aligned}$$