Exercise 1.4.2 ($\sum\limits_{k=0}^n (-1)^k \binom{n}{k} = 0$.)

Question

Show that if $n \geq 1$ then $\sum\limits_{k=0}^n (-1)^k \binom{n}{k} = 0$.

Richard Ganaye · Accepted Answer

If $n\geq 1$, the binomial theorem gives
$$\sum_{k=0}^n (-1)^k \binom{n}{k} = (1 - 1)^n = 0^n = 0.$$
(If $n=0$,
$$\sum_{k=0}^0 (-1)^k \binom{n}{k}  = (-1)^0 \binom{n}{0} = 1,$$and
$(1-1)^0 = 0^0 = 1$.)

Exercise 1.4.2 ($\sum\limits_{k=0}^n (-1)^k \binom{n}{k} = 0$.)

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Exercise 1.4.2 ($\sum\limits_{k=0}^n (-1)^k \binom{n}{k} = 0$.)

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