TAOCP 5.1.3 Exercise 3
Using the binomial theorem, (1-1)^n=\sum_{k=0}^{n}\binom{n}{k}1^{\,n-k}(-1)^k =\sum_{k=0}^{n}\binom{n}{k}(-1)^k .
Exercise 3. [**] [HM25] Evaluate the sum $\sum_{k} \binom{n}{k} (-1)^k$.
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Using the binomial theorem,
$$ (1-1)^n=\sum_{k=0}^{n}\binom{n}{k}1^{,n-k}(-1)^k =\sum_{k=0}^{n}\binom{n}{k}(-1)^k . $$
Hence, for $n>0$,
$$ \sum_{k=0}^{n}\binom{n}{k}(-1)^k=(1-1)^n=0. $$
When $n=0$, the sum consists only of the term $\binom00=1$, so its value is $1$.
Therefore
$$ \boxed{\sum_{k=0}^{n}\binom{n}{k}(-1)^k= \begin{cases} 1, & n=0,\[4pt] 0, & n>0. \end{cases}} $$