Prove $\sum_{k=0}^n {n\choose k} = 2^n$ and $\sum_{k=0}^n (-1)^{k} {n\choose k}=0$ for $n \ge 1$

Question

$\sum_{k=0}^n {n\choose k} = 2^n$ and $\sum_{k=0}^n (-1)^{k} {n\choose k}=0$ how to prove this statement holds for any $n \ge 1$ ?

I am not sure how this released to $(x+y)^n$

Answer 1

\begin{align} (x+y)^n = \sum_{i=0}^n \binom{n}{i} x^i y^{n-k} \end{align} Plugging in $x = y = 1$ gives \begin{align} 2^n = \sum_{i=0}^n \binom{n}{i} \end{align} The second follows from $x = 1$ and $y = -1$, which gives \begin{align} 0 = \sum_{i=0}^n \binom{n}{i} (-1)^i \end{align}