\(% 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 first look at the base case where \(n = 0\), and we have \begin {align*} \RHS = \frac {1}{2^{2 \cdot 0}} \binom {2 \cdot 0}{0} = \frac {1}{2^0} \binom {0}{0} = 1, \end {align*}
and \(\LHS = T_0 = 1\). So the desired statement is satisfied for the base case where \(n = 0\).
Assume the original statement is true for some \(n = k \geq 0\), that \[ T_n = \frac {1}{2^{2n}} \binom {2n}{n}. \]
Consider \(n = k + 1\), we have \begin {align*} T_n & = T_{k + 1} \\ & = \frac {2(k + 1) - 1}{2(k + 1)} T_k \\ & = \frac {2k + 1}{2(k + 1)} \cdot \frac {1}{2^{2k}} \binom {2k}{k} \\ & = \frac {(2k + 1) (2k + 2)}{2(k + 1) 2 (k + 1)} \cdot \frac {1}{2^{2k}} \frac {(2k)!}(k!k!) \\ & = \frac {(2k + 2)!}{(k + 1)! (k + 1)!} \cdot \frac {1}{2^{2k + 2}} \\ & = \frac {1}{2^{2(k + 1)}} \binom {2 (k + 1)}{k + 1}, \end {align*}
which is precisely the statement for \(n = k + 1\).
The original statement is true for \(n = 0\), and given it holds for some \(n = k \geq 0\), it holds for \(n = k + 1\). Hence, by the principle of mathematical induction, the statement \[ T_n = \frac {1}{2^{2n}} \binom {2n}{n} \] holds for all integers \(n \geq 0\), as desired.
By Newton’s binomial theorem, we have \[ (1 - x)^{-\frac {1}{2}} = 1 + \left (- \frac {1}{2}\right ) (-x) + \frac {\left (-\frac {1}{2}\right ) \left (-\frac {3}{2}\right )}{2!} (-x)^2 + \frac {\left (-\frac {1}{2}\right ) \left (-\frac {3}{2}\right ) \left (-\frac {5}{2}\right )}{3!} (-x)^3 + \cdots , \] and notice that the negative signs cancels out, and hence \[ a_n = \frac {\prod _{k = 1}^{n} \frac {2k - 1}{2}}{n!} = \frac {\prod _{k = 1}^{n}(2k - 1)}{2^n n!}. \]
Hence, we note that \begin {align*} \frac {a_r}{a_{r - 1}} & = \frac {\prod _{k = 1}^{r} (2k - 1) / (2^r r!)}{\prod _{k = 1}^{r - 1} (2k - 1) / (2^{r - 1} (r - 1)!)} \\ & = \frac {2r - 1}{2r}, \end {align*}
and hence \[ a_r = \frac {2r - 1}{2r} a_{r - 1}. \]
Note that \(a_0 = 1\) as well. The sequence \(\{a_n\}_{0}^{\infty }\) and \(\{T_n\}_{0}^{\infty }\) have the same initial term \(a_0 = T_0 = 1\), and they have the same same inductive relationship \[ a_n = \frac {2n - 1}{2n} a_{n - 1}, T_n = \frac {2n - 1}{2n} T_{n - 1}. \]
This shows they are the same sequence, hence \[ a_n = T_n \] for all \(n = 0, 1, 2, \cdots \).
By Newton’w binomial theorem, \[ (1 - x)^{-\frac {3}{2}} = 1 + \frac {\left (- \frac {3}{2}\right ) (-x)}{1!} + \frac {\left (- \frac {3}{2}\right ) \left (- \frac {5}{2}\right ) (-x)}{2!} + \cdots , \] and so \[ b_n = \frac {\prod _{k = 1}^{n} \frac {2k + 1}{2}}{n!} = \frac {\prod _{k = 1}^{n} (2k + 1)}{2^n n!}. \]
Notice that \begin {align*} \frac {b_n}{a_n} & = \frac {\prod _{k = 1}^{n} (2k + 1) / (2^n n!)}{\prod _{k = 1}^{n} (2k - 1) / 2^n n!} \\ & = \frac {\prod _{k = 1}^{n} (2k + 1)}{\prod _{k = 1}^{n} (2k - 1)} \\ & = \frac {\prod _{k = 2}^{n + 1} (2k - 1)}{\prod _{k = 1}^{n} (2k - 1)} \\ & = \frac {2(n + 1) - 1}{2 \cdot 1 - 1} \\ & = 2n + 1, \end {align*}
and so \begin {align*} b_n & = (2n + 1) a_n \\ & = (2n + 1) \cdot \frac {1}{2^{2n}} \cdot \binom {2n}{n} \\ & = \frac {2n + 1}{2^{2n}} \binom {2n}{n}. \end {align*}
By the binomial expansion, we have \[ (1 - x)^{-1} = 1 + x + x^2 + x^3 + \cdots , \] and we have \[ (1 - x)^{-\frac {1}{2}} \cdot (1 - x)^{-1} = (1 - x)^{-\frac {3}{2}}. \]
For a particular term in the series expansion for \((1 - x)^{-\frac {3}{2}}\), say \(b_n\), we must have \[ b_n x^n = \sum _{t = 0}^{n} a_t \cdot x^t \cdot 1 \cdot x^{n - t}, \] and hence \[ b_n = \sum _{t = 0}^{n} a_t, \] which gives \[ \frac {2n + 1}{2^{2n}} \binom {2n}{n} = \sum _{r = 0}^{n} \frac {1}{2^{2r}} \binom {2r}{r}, \] exactly as desired.