TITLE: Show $\sum_{k=0}^{n}2^{n-k}\binom{a+k}{k}\frac{a-k}{a+k}=\binom{a+n}{n}$
QUESTION [8 upvotes]: How it can be shown that:

$$\sum_{k=0}^{n}2^{n-k}\binom{a+k}{k}\frac{a-k}{a+k}=\binom{a+n}{n}$$

Where $a \ne 0$

My try:
$$\sum_{k=0}^{n}2^{n-k}\binom{a+k}{k}\frac{a-k}{a+k}=\sum_{k=0}^{n}2^{n-k}\binom{a+k}{a}\frac{a-k}{a+k}$$
$$=\frac{1}{a}\sum_{k=0}^{n}2^{n-k}\binom{a+k-1}{a-1}\left(a-k \right)$$
$$=\sum_{k=0}^{n}2^{n-k}\binom{a+k-1}{a-1}-\frac{1}{a}\color{red}{\sum_{k=0}^{n}2^{n-k}\binom{a+k-1}{a-1}k }$$
On the other hand: 
$$\color{red}{\sum_{k=0}^{n}2^{n-k}\binom{a+k-1}{a-1}k}=\sum_{k=0}^{n}\left(a+k-1\right)2^{n-k}\binom{a+k-2}{k-1}  $$
But computing these expressions takes much time and I think there should be a better way, but I cannot find that way.
Please if it's possible, then do the proof using elementary ways.

REPLY [2 votes]: Here is an algebric proof :
Let $ a $ be a real, let's define the function $ f_{a} $ as follows : $$ \left(\forall x\in\left]-\frac{1}{2},\frac{1}{2}\right[\right),\ f_{a}\left(x\right)=\displaystyle\frac{1-2x}{\left(1-x\right)^{a+1}} $$
know : \begin{aligned} \left(\forall x\in\left]-\frac{1}{2},\frac{1}{2}\right[\right),\ f_{a}\left(x\right)&=\displaystyle\frac{2}{\left(1-x\right)^{a}}-\displaystyle\frac{1}{\left(1-x\right)^{a+1}}\\ &=2\displaystyle\sum_{n=0}^{+\infty}{\left(-1\right)^{n}\displaystyle\binom{-a}{n}x^{n}}-\displaystyle\sum_{n=0}^{+\infty}{\left(-1\right)^{n}\displaystyle\binom{-a-1}{n}x^{n}} \end{aligned}
Observe that forall $ \left(x,n\right)\in\mathbb{R}\times\mathbb{N} $ :
$ \ \ \ \ \ \ \ \ \ \ \left(-1\right)^{n}\binom{-x}{n}=\frac{\left(-1\right)^{n}}{n!}\prod\limits_{k=0}^{n-1}{\left(-x-k\right)}=\frac{1}{n!}\prod\limits_{k=0}^{n-1}{\left(x+k\right)}=\frac{1}{n!}\prod\limits_{k=0}^{n-1}{\left(x+n-1-k\right)}=\binom{x+n-1}{n} $
Thus, \begin{aligned}\left(\forall x\in\left]-\frac{1}{2},\frac{1}{2}\right[\right),\ f_{a}\left(x\right)=\displaystyle\sum_{n=0}^{+\infty}{\left(2\displaystyle\binom{a+n-1}{n}+\displaystyle\binom{a+n}{n}\right)x^{n}}\end{aligned}
Since $ \binom{a+n-1}{n}=\frac{a}{a+n}\binom{a+n}{n} $, we get, \begin{aligned}\left(\forall x\in\left]-\frac{1}{2},\frac{1}{2}\right[\right),\ f_{a}\left(x\right)=\displaystyle\sum_{n=0}^{+\infty}{\displaystyle\frac{a-n}{a+n}\displaystyle\binom{a+n}{n}x^{n}}\end{aligned}
Observe $ \left(\forall x\in\left]-\frac{1}{2},\frac{1}{2}\right[\right),\ \frac{f_{a}\left(x\right)}{1-2x}=\frac{1}{\left(1-x\right)^{a+1}} $, meaning that forall $ x \in\left]-\frac{1}{2},\frac{1}{2}\right[ $, \begin{aligned} \left(\displaystyle\sum_{n=0}^{+\infty}{\displaystyle\frac{a-n}{a+n}\displaystyle\binom{a+n}{n}x^{n}}\right)\left(\displaystyle\sum_{n=0}^{+\infty}{2^{n}x^{n}}\right)&=\displaystyle\sum_{n=0}^{+\infty}{\left(-1\right)^{n}\displaystyle\binom{-a-1}{n}x^{n}} \\ \iff \displaystyle\sum_{n=0}^{+\infty}{\left(\displaystyle\sum_{k=0}^{n}{2^{n-k}\displaystyle\binom{a+k}{k}\displaystyle\frac{a-k}{a+k}}\right)x^{n}}&=\displaystyle\sum_{n=0}^{+\infty}{\displaystyle\binom{a+n}{n}x^{n}}\end{aligned}
Which leads to the following : $$ \left(\forall n\in\mathbb{N}\right),\ \displaystyle\sum_{k=0}^{n}{2^{n-k}\displaystyle\binom{a+k}{k}\displaystyle\frac{a-k}{a+k}}=\displaystyle\binom{a+n}{n} $$