Determine (without doing heavy calculations) the value of

$$
\binom{6}{0}-\binom{6}{1}+\binom{6}{2}-\binom{6}{3}+\binom{6}{4}-\binom{6}{5}+\binom{6}{6}
$$

Generalize by replacing 6 with $n$.
By taking $n=6, a=1$ and $b=-1$ in the proposition, since $(-1)^{k}$ is 1 if $k$ is even and -1 if $k$ is odd, we obtain that the first expression is $(1-1)^{6}=0$. More generally, we obtain for any $n$ that

$$
\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3}+\ldots+(-1)^{n-1}\binom{n}{n-1}+(-1)^{n}\binom{n}{n}=0
$$

We can summarize the result of the previous exercise by saying that the alternating sum of the coefficients of Pascal's triangle is 0.

## Inequalities

This section is an introduction to inequalities, which represent a significant class of exercises in olympiads. It always involves showing that a certain quantity depending on one or more variables is greater than another quantity, when the variables in question take values in a certain set (which can be $\mathbb{R}$, or $\mathbb{R}^{+}$, or something else). Often, it is also necessary to determine the equality cases, that is, the values of the variables for which these two quantities are equal. To prove inequalities, one is allowed to use certain known inequalities.

The simplest and most well-known inequality is the one that states a square is always non-negative:

Proposition 116. For all $x \in \mathbb{R}$, we have $x^{2} \geq 0$. Moreover, equality holds if and only if $x=0$.

Despite its seemingly harmless appearance, this inequality is very useful and allows us to prove many things.