Binomial Theorem: does $\sum\limits_{i=0}^{n}{{n}\choose{i}} = \dfrac{1}{2}\sum\limits_{i=0}^{n+1}{{n+1}\choose{i}}$? I am reading through an argument that I found somewhere and I see that it assumes that:
$$\sum\limits_{i=0}^{n}{{n}\choose{i}} = \dfrac{1}{2}\sum\limits_{i=0}^{n+1}{{n+1}\choose{i}}$$
Intuitively, this seems wrong.  I'm assuming it is a typo.  
Here's my reasoning:
$$\sum\limits_{i=0}^{n+1}{{n+1}\choose{i}} = (n+1)\sum\limits_{i=0}^{n}\frac{1}{n+1-i}{{n}\choose{i}} + {{n+1}\choose{n+1}} = (n+1)\sum\limits_{i=0}^{n}\frac{1}{n+1-i}{{n}\choose{i}} + 1$$
So that:
$$\sum\limits_{i=0}^{n}\frac{n+1}{n+1-i}{{n}\choose{i}} = \sum\limits_{i=0}^{n+1}{{n+1}\choose{i}} - 1$$
How do I finish my argument to show the assumption is incorrect?  Is it a typo or is bad reasoning?  Or am I incorrect and the result is valid?
 A: Using the recurrence for Pascal's Triangle and the fact that $\binom{n}{k}=0$ when $k\lt0$ or $k\gt n$, we get
$$
\begin{align}
\sum_{i=0}^{n+1}\binom{n+1}{i}
&=\sum_{i=0}^{n+1}\left[\binom{n}{i}+\binom{n}{i-1}\right]\\
&=\sum_{i=0}^{n+1}\binom{n}{i}+\sum_{i=-1}^n\binom{n}{i}\\
&=\sum_{i=0}^n\binom{n}{i}+\sum_{i=0}^n\binom{n}{i}\\
&=2\sum_{i=0}^n\binom{n}{i}
\end{align}
$$
A: $2^n=(1+1)^n= \frac{1}{2}2^{n+1}=\frac{1}{2}(1+1)^{n+1}$ $\Rightarrow$ $\displaystyle \sum_{i=0}^n \binom{n}{i}= \frac{1}{2}\displaystyle \sum_{i=0}^{n+1}\binom{n+1}{i}$
A: This is not an answer but it is too long for a comment.
I checked the expressions and played with numerical values and what I observed is that $$1+\sum\limits_{i=0}^{n}{{n}\choose{i}} = \dfrac{1}{2}\sum\limits_{i=0}^{n+1}{{n+1}\choose{i}}$$ I am too bad in this area to claim that this proves anything but the above equality seems to hold for any $n$ I tried.
Afterwards, this looks to me to be normal since $$\sum\limits_{i=0}^{n}{{n}\choose{i}}=2^n-1$$ $$\sum\limits_{i=0}^{n+1}{{n+1}\choose{i}}=2^{n+1}$$ However, taking into account the other answers to this post, I may be totally wrong.
