For all $n \in \mathbb { N } ^ { * }$ and all $k \in \llbracket 0 , n \rrbracket$, we set $x _ { n , k } = - \sqrt { n } + \frac { 2 k } { \sqrt { n } }$. The function $B _ { n } : \mathbb { R } \rightarrow \mathbb { R }$ is defined by $$\forall x \in ] - \infty , - \sqrt { n } - \frac { 1 } { \sqrt { n } } [ , \quad B _ { n } ( x ) = 0$$ $$\forall k \in \llbracket 0 , n \rrbracket , \forall x \in \left[ x _ { n , k } - \frac { 1 } { \sqrt { n } } , x _ { n , k } + \frac { 1 } { \sqrt { n } } [ , \right. \quad B _ { n } ( x ) = \frac { \sqrt { n } } { 2 } \binom { n } { k } \frac { 1 } { 2 ^ { n } }$$ $$\forall x \in \left[ \sqrt { n } + \frac { 1 } { \sqrt { n } } , + \infty [ , \right. \quad B _ { n } ( x ) = 0.$$ For all $n \in \mathbb { N } ^ { * }$, show that $B _ { n }$ is a decreasing function on $\mathbb { R } ^ { + }$. One may distinguish according to whether $n$ is even or odd.
For all $n \in \mathbb { N } ^ { * }$ and all $k \in \llbracket 0 , n \rrbracket$, we set $x _ { n , k } = - \sqrt { n } + \frac { 2 k } { \sqrt { n } }$. The function $B _ { n } : \mathbb { R } \rightarrow \mathbb { R }$ is defined by
$$\forall x \in ] - \infty , - \sqrt { n } - \frac { 1 } { \sqrt { n } } [ , \quad B _ { n } ( x ) = 0$$
$$\forall k \in \llbracket 0 , n \rrbracket , \forall x \in \left[ x _ { n , k } - \frac { 1 } { \sqrt { n } } , x _ { n , k } + \frac { 1 } { \sqrt { n } } [ , \right. \quad B _ { n } ( x ) = \frac { \sqrt { n } } { 2 } \binom { n } { k } \frac { 1 } { 2 ^ { n } }$$
$$\forall x \in \left[ \sqrt { n } + \frac { 1 } { \sqrt { n } } , + \infty [ , \right. \quad B _ { n } ( x ) = 0.$$
For all $n \in \mathbb { N } ^ { * }$, show that $B _ { n }$ is a decreasing function on $\mathbb { R } ^ { + }$. One may distinguish according to whether $n$ is even or odd.