Use the binomial theorem to show that for any positive integer $n$, $\displaystyle\sum_{i=0}^{n} {n \choose i} = 2^n$.

Question

Can somebody check to see if this is good enough just to show? It's very simple but the question doesn't say prove or anything like that.

So the binomial theorem states that $(x+y)^n=\displaystyle\sum_{r=0}^{n} {n \choose r}x^{n-r}y^r$

Let $x=1, y=1$.

Then $2^n=\displaystyle\sum_{r=0}^{n} {n \choose r}*1^{n-r}1^r$, which reduces to $2^n=\displaystyle\sum_{r=0}^{n} {n \choose r}$. Tada.

Good enough?

Answer 1

This is a perfectly valid proof, you're done.

Answer 2

Your proof is correct. If I might add an interpretation to the above result, it says that the total number of subsets from a set $A$ of $n$ items is $2^n$. You might run across that result in set theory or elementary real analysis.

Answer 3

in the binomial theorem take a=b=1