Finding the summation of n choose r

Question

I am just trying to understand how to find the summation of a basic combination, in order to do the ones on my assignment, and would be grateful if someone could take me step by step on how to get the summation of: $$ \sum\limits_{k=0}^n {n\choose k} $$ I believe that the Binomial Theorem should be used, but I am unsure of how/ what to do?

I would really appreciate any help. Thanks in advance.

Answer 1

From binomial theorem $$(1+t)^n=\sum_{k=0}^n\binom{n}{k}t^{k}$$ for $t=1$ we get that $$(1+1)^n=2^n=\sum_{k=0}^n\binom{n}{k}$$

Answer 2

$$ (x + y)^n = \sum_{k = 0}^n{n \choose k}x^{n - k}y^k ~ \rightarrow ~ (1 + y)^n = \sum_{k = 0}^n {n \choose k}y^{k} ~ \rightarrow ~ 2^n = \sum_{k = 0}^n{n \choose k} $$

Answer 3

you should use the Binomial Theorem with $x=y=1$.

you get:

$$(x+y)^n = \sum\limits_{k=0}^n {n\choose k}x^n y^{n-k} \Rightarrow (1+1)^n = \sum\limits_{k=0}^n {n\choose k}1^n 1^{n-k} \Rightarrow 2^n = \sum\limits_{k=0}^n {n\choose k} $$

Answer 4

Alternative by induction:

Define $\binom{n}{k}:=0$ if $k\notin\left\{ 0,\ldots,n\right\} $ and $s_{n}=\sum_{k\in\mathbb{Z}}\binom{n}{k}$. Then $s_{0}=1$ and $s_{n+1}=\sum_{k\in\mathbb{Z}}\binom{n+1}{k}=\sum_{k\in\mathbb{Z}}\binom{n}{k-1}+\binom{n}{k}=2s_{n}$.

This makes it easy to prove by induction that $s_{n}=2^{n}$.