Prove with combinatorial arguments this equation

Question

Prove with combinatorial arguments, that, $\forall n \in \mathbb{N}$.

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

Answer 1

Using mathematical induction, we have:

$$n = 1\,\, \to \,\,\left( {\begin{array}{*{20}{c}} 1\\0 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right) = 1 - 1 = 0$$

Now we must show that if $$\sum\limits_{k = 0}^n {{{\left( { - 1} \right)}^k}\left( {\begin{array}{*{20}{c}} n\\k \end{array}} \right)} = 0$$ then $$\sum\limits_{k = 0}^{n + 1} {{{\left( { - 1} \right)}^k}\left( {\begin{array}{*{20}{c}} {n + 1}\\k \end{array}} \right)} = 0$$

Using the identity $$ \left( {\begin{array}{*{20}{c}} {n + 1}\\k\end{array}} \right) = \left( {\begin{array}{*{20}{c}} n\\k\end{array}} \right) + \left( {\begin{array}{*{20}{c}} n\\{k - 1}\end{array}} \right) $$ we have, $$\sum\limits_{k = 0}^{n + 1} {{{\left( { - 1} \right)}^k}\left( {\begin{array}{*{20}{c}} {n + 1}\\k\end{array}} \right)} = \sum\limits_{k = 0}^{n + 1} {{{\left( { - 1} \right)}^k}\left( {\begin{array}{*{20}{c}} n\\k\end{array}} \right)} + \sum\limits_{k = 0}^{n + 1} {{{\left( { - 1} \right)}^k}\left( {\begin{array}{*{20}{c}} n\\{k - 1}\end{array}} \right)} = \sum\limits_{k = 0}^n {{{\left( { - 1} \right)}^k}\left( {\begin{array}{*{20}{c}} n\\k\end{array}} \right)} + \sum\limits_{k = - 1}^n {{{\left( { - 1} \right)}^{k + 1}}\left( {\begin{array}{*{20}{c}} n\\k\end{array}} \right)} = 0 - 0 = 0$$ which completes the proof. Note that we have used $$\left( {\begin{array}{*{20}{c}}n\\{n + 1}\end{array}} \right) = \left( {\begin{array}{*{20}{c}} n\\{ - 1}\end{array}} \right) = 0$$ which are the direct consequences of the identity $\frac{1}{(-1)!}=0$