Prove by induction that for all $n \ge 0$:
$${n \choose 0} + {n \choose 1} + ... + {n \choose n} = 2^n.$$
In the inductive step, use Pascal’s identity, which is:
$${n+1 \choose k} = {n \choose k-1} + {n \choose k}.$$
I can only prove it using the binomial theorem, not induction.
For basic step n=0:
$\binom{0}{0}=\frac{0!}{0!0!}=2^0$
For induction step:
Let k be an integer such that $0\lt{k}$ and for all L, $0\le{L}\le{k}$ where $L\in{I}$, the formula stand true.
Then:
$$\binom{k}{0}+\binom{k}{1}+...+\binom{k}{k}=2^k$$
Now as can be illustrated easily $\binom{k}{0}=\binom{k+1}{0}$ and $\binom{k}{k}=\binom{k+1}{k+1}$.
Now by using Pascal's identity,
$$\begin{align}\binom{k+1}{0}+\binom{k+1}{1}+\binom{k+1}{2}+...+\binom{k+1}{k}+\binom{k+1}{k+1}\\=\binom{k+1}{0}+\binom{k}{0}+\binom{k}{1}+\binom{k}{1}+\binom{k}{2}+...+\binom{k}{k-1}+\binom{k}{k}+\binom{k+1}{k+1}\\=\binom{k}{0}+\binom{k}{0}+\binom{k}{1}+\binom{k}{1}+\binom{k}{2}+...+\binom{k}{k-1}+\binom{k}{k}+\binom{k}{k}\\=2*{\sum_{i=0}^k\binom{k}{i}}\\=2*2^k\\=2^{k+1}\end{align}$$
As the formula is also true for $k+1$ hence by second principle of finite induction this formula is valid for all integers greator than or equal to $0$.
Well, here it is : $$ \sum_{k=0}^{n+1} {n+1\choose k} = {n+1\choose 0 } + {n+1\choose 1 } + \ldots + {n+1\choose n } + {n+1\choose n+1} $$ $$ = 1 + {n\choose 0} + {n\choose 1} + {n\choose 1} + {n\choose 2 } + \ldots + {n\choose n-1} + {n\choose n} + 1 $$ $$ = {n\choose 0 } + {n\choose 0 } + {n\choose 1} + {n\choose 1 } + \ldots + {n\choose n-1} + {n\choose n-1} + {n\choose n} + {n\choose n} $$ $$ = 2 \times \sum_{k=0}^n {n\choose k } $$ Now induct!
In case anyone wants an alternative, from the binomial theorem,
$$(x+y)^n=\sum_{i=0}^{n}{n\choose i}x^{n-i}y^i$$
If we let $x=1$ and $y=1$, then $\sum_{i=0}^{n}{n\choose i}=2^n$.
Since someone decided to revive this 6 year old question, you can also prove this using combinatorics
Take n elements and count how many ways there are to put these two elements into 2 different containers (A and B)
a) Every element can take two states: it's either in A or in B. By one of the first theorems in combinatorics (the one about separating a task into steps), this gives n steps with 2 options at each step, so we get $2^n$
b) We can also think of adding up all the ways in which we can have k elements in A and n-k elements in B, for $0 \le k \le n$. For each k, this number is: "choose k elements that will go into A". Then the other n-k elements automatically go to B. So for each k this number is just ${n \choose k}$. Now, we add this for all k reviously mentioned to get $\sum_{k = 0}^n {n \choose k}$
And thus, $2^n = \sum_{k = 0}^n {n \choose k}$
