TITLE: Sum from $0$ to $n$ of $ n \choose i $?
QUESTION [6 upvotes]: 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

REPLY [4 votes]: 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.