TITLE: How to prove $f(n)=\sum_{k=0}^n\binom{n+k}{k}\left(\frac{1}{2}\right)^k=2^n$ without using the induction method?
QUESTION [5 upvotes]: The problem is to prove the following statement for all natural numbers.
$$f(n)=\sum_{k=0}^n\binom{n+k}{k}\left(\frac{1}{2}\right)^k=2^n$$
I already proved this using mathematical induction.
But, my instinct is telling me that there is some kind of combinatorial/algebraic method that can be used to solve this easily. (Unless my instinct is wrong.)
Can anyone give me hint on how to do this?

REPLY [4 votes]: Here is a generating function proof that avoids (or hides) the induction.
$$
\begin{align}
f(x)
&=\sum_{n=0}^\infty\sum_{k=0}^n\binom{n+k}{k}\left(\frac12\right)^{n+k}x^n\tag{1}\\
&=\sum_{n=0}^\infty\sum_{k=0}^n\left[\binom{n+k-1}{k}+\binom{n+k-1}{k-1}\right]\left(\frac12\right)^{n+k}x^n\tag{2}\\
&=\sum_{n=0}^\infty\sum_{k=0}^n\binom{n+k-1}{k}\left(\frac12\right)^{n+k}x^n+\sum_{n=0}^\infty\sum_{k=0}^{n-1}\binom{n+k}{k}\left(\frac12\right)^{n+k+1}x^n\tag{3}\\
&=\frac12+\sum_{n=0}^\infty\sum_{k=0}^{n-1}\binom{n+k-1}{k}\left(\frac12\right)^{n+k}x^n+\sum_{n=0}^\infty\sum_{k=0}^n\binom{n+k}{k}\left(\frac12\right)^{n+k+1}x^n\tag{4}\\
&=1+\sum_{n=0}^\infty\sum_{k=0}^{n-1}\binom{n+k-1}{k}\left(\frac12\right)^{n+k-1}x^n\tag{5}\\
&=1+\sum_{n=0}^\infty\sum_{k=0}^n\binom{n+k}{k}\left(\frac12\right)^{n+k}x^{n+1}\tag{6}\\[9pt]
&=1+x\,f(x)\tag{7}
\end{align}
$$
Explanation:
$(2)$: Pascal's Triangle relation
$(3)$: substitute $k\mapsto k+1$ in the right-hand sum
$(4)$: use $\binom{2n-1}{n}\left(\frac12\right)^{2n}=\binom{2n}{n}\left(\frac12\right)^{2n+1}$ for $n\ge1$ to move $x^n$ term from the left to the right sum
$(5)$: twice $(4)$ minus $(1)$
$(6)$: substitute $n\mapsto n+1$
$(7)$: compare $(1)$ and $(6)$
Note that $(7)$ implies that
$$
\begin{align}
f(x)
&=\frac1{1-x}\\
&=\sum_{n=0}^\infty x^n\tag{8}
\end{align}
$$
Comparing the coefficients of $(1)$ and $(8)$ yields
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{k=0}^n\binom{n+k}{k}\left(\frac12\right)^{n+k}=1}\tag{9}
$$

Postscript
I think it is interesting to note that
$$
\begin{align}
\sum_{k=0}^\infty\binom{n+k}{k}x^{n+k}
&=\sum_{k=0}^\infty(-1)^k\binom{-n-1}{k}x^{n+k}\\
&=\frac{x^n}{(1-x)^{n+1}}\tag{10}
\end{align}
$$
Setting $x=\frac12$ in $(10)$ yields
$$
\sum_{k=0}^\infty\binom{n+k}{k}\left(\frac12\right)^{n+k}=2\tag{11}
$$
Thus, exactly half of the sum resides in the first $n+1$ terms.