# Sum of combinations of the n by consecutive k

In a book, I found that the sum of combinations: $\binom{n}{k} + \binom{n}{k+1} +\cdots+ \binom{n}{n}$, where k starts from 0, equals $2^n$. It is possible to express this statement via sum: $2 + \sum_{k=1}^{n-1}{n!\over k!(n-k)!} = 2^n$, using binomial theorem. I tried several small n values, such as 4, 6 and others, the statement looks correct. But I can’t find a regularity, respectively I can’t understand, how to prove the truth of the statement?

#### Solutions Collecting From Web of "Sum of combinations of the n by consecutive k"

$$(x+y)^n=\sum_{k=0}^n \binom{n}{k} x^k y^{n-k}$$

For $x=1$ and $y=1$:

$$2^n=\sum_{k=0}^n \binom{n}{k}=\binom{n}{0}+\binom{n}{1}+ \dots +\binom{n}{n-1}+\binom{n}{n}$$