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
$$S ( x ) = \sum _ { k = 0 } ^ { n } \left| x - \frac { k } { n } \right| \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k } .$$
We denote
\begin{itemize}
\item $V$ the set of integers $k \in \{ 0 , \ldots , n \}$ such that $\left| x - \frac { k } { n } \right| \leqslant \frac { 1 } { \sqrt { n } }$,
\item $W$ the set of integers $k \in \{ 0 , \ldots , n \}$ such that $\left| x - \frac { k } { n } \right| > \frac { 1 } { \sqrt { n } }$, and we set
\end{itemize}
$$S _ { V } ( x ) = \sum _ { k \in V } \left| x - \frac { k } { n } \right| \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k } \quad \text { and } \quad S _ { W } ( x ) = \sum _ { k \in W } \left| x - \frac { k } { n } \right| \binom { n } { k } x ^ { k } ( 1 - x ) ^ { n - k } .$$
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 $S ( x ) \leqslant \frac { 5 } { 4 \sqrt { n } }$.