Determine the value of summation form. Determine the value of
$\displaystyle \sum_{k=n}^{2n} \binom{k} {n} 2^{-k}$ for $n \geq 1$
My attempt:
I have asked this question before in another math forum but it leads no solution. My friend suggested me to search for something like series-k. I also looked at my textbook in combinatoric section (which apparently leads to Binomial Theorem), but really I don't have any idea for this one. 
Could you help me? 
 A: Let's start with the famous identity known from Pascal's triangle,
$$\binom{k}{n}=\binom{k-1}{n-1}+\binom{k-1}{n}.$$ It's valid for all $k$ and $n$, if we assume $\binom{k}{n}=0$ for $n>k$ or $n<0$. So we can define our sum as
$$s_n=\sum_{k\le2n}\binom{k}{n}2^{-k}=\sum_{k\le2n}\binom{k-1}{n-1}2^{-k}+\sum_{k\le2n}\binom{k-1}{n}2^{-k}$$ Obviously, we can rewrite the RHS as
$$\frac12\sum_{k\le2n}\binom{k-1}{n-1}2^{-k+1}+\frac12\sum_{k\le2n}\binom{k-1}{n}2^{-k+1},$$ and with a translation of the summation index, this becomes
$$\frac12\sum_{k\le2n-1}\binom{k}{n-1}2^{-k}+\frac12\sum_{k\le2n-1}\binom{k}{n}2^{-k}.$$ The sums on the RHS can be expressed by $s_{n-1}$ and $s_n$, so we obtain 
$$\frac12\,s_{n-1}+\binom{2n-1}{n-1}2^{-2n}+\frac12\,s_n-\binom{2n}{n}2^{-2n-1}=\frac12\,s_{n-1}+\frac12\,s_n$$ because of $$\binom{2n}{n}=\binom{2n-1}{n-1}+\binom{2n-1}{n}=2\binom{2n-1}{n-1}.$$ But this implies $s_n=s_{n-1}$, and the final result is $$s_n=s_1=\binom{1}{1}2^{-1}+\binom{2}{1}2^{-2}=1.$$
A: This answer is based upon the Lagrange inversion formula. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write for instance
\begin{align*}
[z^n]\frac{1}{1-2z}=[z^n]\sum_{j=0}^\infty 2^jz^j=2^n
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{k=n}^{2n}\binom{k}{n}2^{-k}}
&=\sum_{k=0}^n\binom{n+k}{k}2^{-n-k}\tag{1}\\
&=\frac{1}{2^n}\sum_{k=0}^n\binom{-n-1}{k}\left(-\frac{1}{2}\right)^k\tag{2}\\
&=\frac{1}{2^n}[z^n]\frac{1}{\left(1-\frac{z}{2}\right)^{n+1}(1-z)}\tag{3}\\
&=\frac{1}{2^n}[z^n]\left.\left(\frac{1}{\left(1-\frac{w}{2}\right)(1-w)}\cdot\frac{1}{1-\frac{w}{2}\cdot\frac{1}{1-\frac{w}{2}}}\right)\right|_{w=\frac{z}{1-\frac{w}{2}}}\tag{4}\\
&=\frac{1}{2^n}[z^n]\left.\frac{1}{(1-w)^2}\right|_{w=1-\sqrt{1-2z}}\tag{5}\\
&=\frac{1}{2^n}[z^n]\frac{1}{1-2z}\\
&\color{blue}{=1}
\end{align*}

Comment:


*

*In (1) we shift the index $k$ to start with $k=0$.

*In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$ and $\binom{p}{q}=\binom{p}{p-q}$.

*In (3) we observe the sum is the coefficient of the convolution of two series
\begin{align*}
[z^n]\left(\sum_{k}a_kz^k\right)\left(\sum_{l}b_lz^l\right)=[z^n]\sum_{N}\left(\sum_{k=0}^Na_k b_{N-k}\right)z^N
=\sum_{k=0}^na_k b_{n-k}
\end{align*}
Here with $a_k=\binom{-n-1}{k}\left(-\frac{1}{2}\right)^k$ and $b_k=1$.

*In (4) we use the Lagrange Inversion Formula in the form G6 stated in R. Sprugnolis (etal) paper Lagrange Inversion: when and how. 
\begin{align*}
[z^n]F(z)\Phi(z)^n=[z^n]\left.\frac{F(w)}{1-z\Phi'(w)}\right|_{w=z\Phi(w)}
\end{align*}
Here with $\Phi(z)=\frac{1}{1-\frac{z}{2}}$ and $F(z)=\frac{1}{\left(1-\frac{z}{2}\right)(1-z)}$. It follows since $w=z\Phi(w)$:
\begin{align*}
z\Phi^{\prime}(w)=\frac{z}{2}\cdot\frac{1}{\left(1-\frac{w}{2}\right)^2}
=\frac{z}{2}\Phi^2(w)=\frac{w}{2}\Phi(w)=\frac{w}{2}\cdot\frac{1}{1-\frac{w}{2}}
\end{align*}


*

*In (5) we simplify the expression and we select the solution $w=w(z)$ from $w=\frac{z}{1-\frac{w}{2}}$ which can be expanded in a power series.

