Sum from $0$ to $n$ of $ n \choose i $? 
Possible Duplicate:
Algebraic Proof that $\sum\limits_{i=0}^n \binom{n}{i}=2^n$
Evaluation $\sum\limits_{k=0}^n \binom{n}{k}$

Is there a simple proof for this equality:
$$\sum_0^n  {n \choose i}  = 2^n$$
thanks and sorry I forgot the basics
 A: The standard combinatorial proof is that


*

*The LHS counts the number of ways to choose $0$, $1$, $2, \ldots , $ or $ n$ things from a total of $n$ objects.

*The RHS counts the number of ways to go through each of $n$ objects and mark them as "choose" or "don't choose".


With a little thought, these are equal.
An algebraic proof has been posted by Siminore.
A: Recall the relation $\displaystyle{n+1\choose i}={n\choose i}+{n\choose i-1}$, valid for every $1\leqslant i\leqslant n$. Hence,
$$
\sum_{i=0}^{n+1}{n+1\choose i}=1+\sum_{i=1}^n\left[{n\choose i}+{n\choose i-1}\right]+1,
$$
that is,
$$
\sum_{i=0}^{n+1}{n+1\choose i}=1+\sum_{i=1}^n{n\choose i}+\sum_{i=0}^{n-1}{n\choose i}+1=2\sum_{i=0}^n{n\choose i}.
$$
The initial value $\displaystyle\sum\limits_{i=0}^0{0\choose i}=1$ completes the recursion.
A: Here's a variation on the theme of Didier's answer.
Each number in Pascal's triangle gets added twice to the row below it.
The first $1$ below gets added to the next row to get the $1$ at the end, and also gets added to the next row to contribute to the $9$.  Then the $8$ gets added to the next row to contribute to the $9$, and also gets added to the next row to contribute to the $36$.  And so on.
$$
\begin{array}{ccccccccccccccccccc}
& 1 & & 8 & & 28 & & 56 & & 70 & & 56 & & 28 & & 7 & & 1 \\  \\
1 & & 9 & & 36 & & 84 & & 126 & & 126 & & 84 & & 36 & & 9 & & 1
\end{array}
$$
Since each number is added twice to the next row, the sum of the numbers in the next row is twice as big.
A: $$(1+1)^n = \sum_{i=0}^n \begin{pmatrix} n \\ i \end{pmatrix}1^i 1^{n-i}.$$ But I do not know if this is as simple as you wish.
A: There are many quite simple proofs.
One of them is application of well known Newton binomial theorem:
$$
\sum_0^n {n \choose i} = \sum_0^n {n \choose i} 1^i 1^{n-i} = (1+1)^n = 2^n.
$$
One can also prove this by combinatorial argument. Observe that ${n \choose i}$ is the number of subsets of cardinality $i$ of a set of cardinality $n$. Then $\sum_0^n {n \choose i}$ is the number of all subsets of cardinality $0, 1, 2, \dots, n$ of a set of cardinality $n$. Hence the sum counts all subsets of a $n$-set. But we know that thare are $2^n$ subsets of such set.
