\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
We notice that \begin {align*} \LHS & = r \binom {2n}{r} \\ & = r \cdot \frac {(2n)!}{r! (2n - r)!} \\ & = \frac {(2n)!}{(r - 1)! (2n - r)!}, \end {align*}
and \begin {align*} \RHS & = (2n + 1 - r) \binom {2n}{2n + 1 - r} \\ & = (2n + 1 - r) \cdot \frac {(2n)!}{(r - 1)! (2n + 1 - r)!} \\ & = \frac {(2n)!}{(r - 1)! (2n - r)!}. \end {align*}
Hence, \[ r \binom {2n}{r} = (2n + 1 - r) \binom {2n}{2n + 1 - r} \] as desired.
Summing this from \(r = n + 1\) to \(2n\), we have \begin {align*} \sum _{r = n + 1}^{2n} r \binom {2n}{r} & = \sum _{r = n + 1}^{2n} (2n + 1 - r) \binom {2n}{2n + 1 - r} \\ & = \sum _{r = 1}^{n} (2n + 1 - (2n + 1 - r)) \binom {2n}{2n + 1 - (2n + 1 - r)} \\ & = \sum _{r = 1}^{n} r \binom {2n}{r}, \end {align*}
and hence \begin {align*} \sum _{r = 0}^{2n} r \binom {2n}{r} & = \sum _{r = 1}^{2n} r \binom {2n}{r} \\ & = \sum _{r = 1}^{n} r \binom {2n}{r} + \sum _{r = n + 1}^{2n} r \binom {2n}{r} \\ & = \sum _{r = n + 1}^{2n} r \binom {2n}{r} + \sum _{r = n + 1}^{2n} r \binom {2n}{r} \\ & = 2 \sum _{r = n + 1}^{2n} r \binom {2n}{r}, \end {align*}
as desired.
For \(n + 1 \leq x \leq 2n\), we have \[ \Prob (X = x) = 2 \cdot \frac {\binom {2n}{x}}{2^{2n}}. \]
For \(x = n\), we have \[ \Prob (X = x) = \frac {\binom {2n}{n}}{2^{2n}}. \]
We have \(n \leq X \leq 2n\), and hence \begin {align*} \Expt (X) & = \sum _{x = n}^{2n} x \Prob (X = x) \\ & = \frac {n \binom {2n}{n}}{2^{2n}} + \frac {2}{2^{2n}} \sum _{x = n + 1}^{2n} x \binom {2n}{x} \\ & = \frac {n \binom {2n}{n}}{2^{2n}} + 2^{-2n} \sum _{r = 0}^{2n} r \binom {2n}{r} \\ & = \frac {n \binom {2n}{n}}{2^{2n}} + 2^{-2n} (2n) 2^{2n - 1} \\ & = n + \frac {n \binom {2n}{n}}{2^{2n}} \\ & = n \left (1 + \frac {1}{2^{2n}} \binom {2n}{n}\right ) \end {align*}
as desired.
First, we have that \[ \frac {1}{2^{2n}} \binom {2n}{n} > 0 \] for all positive integers \(n\).
Taking the ratio of two consecutive terms, we have \begin {align*} \frac {\frac {1}{2^{2n}} \binom {2n}{n}}{\frac {1}{2^{2(n + 1)}} \binom {2(n + 1)}{n + 1}} & = \frac {2^{2n + 2} \frac {(2n)!}{n! n!}}{2^{2n} \frac {(2n + 2)!}{(n + 1)! (n + 1)!}} \\ & = 4 \cdot \frac {(n + 1)^2}{(2n + 2) (2n + 1)}. \end {align*}
We have that the following are equivalent: \begin {align*} \frac {1}{2^{2n}} \binom {2n}{n} & > \frac {1}{2^{2(n + 1)}} \binom {2(n + 1)}{n + 1} \\ \frac {\frac {1}{2^{2n}} \binom {2n}{n}}{\frac {1}{2^{2(n + 1)}} \binom {2(n + 1)}{n + 1}} & > 1 \\ \frac {4 (n + 1)^2}{(2n + 2) (2n + 1)} & > 1 \\ 4n^2 + 8n + 4 & > 4n^2 + 6n + 2 \\ 2n + 2 & > 0 \end {align*}
and this obviously true for all positive integers \(n\).
This means that \(\frac {1}{2^{2n}} \binom {2n}{n}\) decreases as \(n\) increases.