Verifying how the sum $\sum_{k=0}^n (-1)^k\binom{n}{k}$ works

Question

I'm a little bit confused. This is a very easy question but I'm solving lots of exercises and I want to be crystal clear about that.

I want to verify how this sum works.

$$\sum_{k=0}^n (-1)^k\binom{n}{k}$$

Let's take $n=2$. So we will have $$(-1)^0\binom{2}{0}+(-1)^1\binom{2}{1}+(-1)^2\binom{2}{2}=2-2=0$$ Is that true ?

Thank you

Answer 1

Hints:

Find the expansion of $\left(1-x\right)^n$ and substitute $x=1$

Spoiler:

$\quad\left(1-x\right)^n \\ = \sum_{k=0}^n \begin{pmatrix} n\\k \end{pmatrix}1^{n-k}\left(-x\right)^k \\ =\sum_{k=0}^n \left(-1\right)^k x^k\begin{pmatrix} n\\k \end{pmatrix}\\ \text{Substitute both side by $x=1$ and we get}\\ \left(1-1\right)^n=0=\sum_{k=0}^n \left(-1\right)^k\begin{pmatrix} n\\ k \end{pmatrix}$