Exercise 2.1.3

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

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ewcommand{\deb}{\begin{eqnarray*}}

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ewcommand{\ssi} {si et seulement si }

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ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

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ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Given positive integers $n$ and $r$ with $1\leq r \leq n$, let $\binom{n}{r}$ be the number of ways of choosing $r$ elements from a set with $n$ elements. Recall that $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.
\begin{enumerate}
\item[(a)] Show that the polynomial $\sigma_r$ is a sum of $\binom{n}{r}$ terms.
\item[(b)] Show that $\sigma_r(-\alpha,\ldots,-\alpha) = (-1)^r \binom{n}{r} \alpha^r$.
\item[(c)] Let $f = (x+\alpha)^n$. Use part (b) and Corollary 2.1.5 to prove that
$$(x+\alpha)^n = \sum_{r=0}^n \binom{n}{r} \alpha^r x^{n-r},$$
where $\binom{n}{0}=1$. This shows that the binomial theorem follows from Corollary 2.1.5.
\end{enumerate}

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
\begin{enumerate}
\item[(a)]
The number of terms in 
\begin{align}
\sigma_r = \sum_{1 \leq i_1 <  \cdots <i_r\leq n}x_{i_1}x_{i_2}\cdots x_{i_r} \label{eq2.1.3:1}
\end{align}
is the number of strictly increasing sequences $(i_1,i_2,\cdots,i_r)$ in the integer interval $\gcro1,n\dcro$. It is equal to the number of subsets with $r$ elements in the set $\gcro1,n\dcro$ with $n$ elements. Thus it is equal to $\binom{n}{r}$.

\item[(b)] Evaluating \eqref{eq2.1.3:1} with $x_1=x_2 = x_n = -\alpha$, we obtain
\begin{align*}
\sigma_r (-\alpha, \cdots,-\alpha)&= \sum_{1 \leq i_1 <  \cdots <i_r\leq n}(-\alpha)^r\
&=(-1)^r \binom{n}{r} \alpha^r.
\end{align*}

\item[(c)]
By Corollary 2.1.5, with the substitution $x_1 = -\alpha, x_2 = -\alpha, \cdots, x_n = - \alpha$,
$$f = (x+\alpha)^n = x^n +a_1 x^{n-1}+\cdots +a_n,$$ where 
\begin{align*}
a_r &= (-1)^r \sigma_r(-\alpha,\cdots,-\alpha)\
&=\binom{n}{r} \alpha^r.
\end{align*}

Consequently, 
$$(x+ \alpha)^n = \sum_{i=1}^n \binom{n}{r} \alpha^r x^{n-r}.$$

With the substitution $x = \beta , \ \beta \in F$, we obtain the binomial formula

$$( \alpha+ \beta)^n = \sum_{i=1}^n \binom{n}{r} \alpha^r \beta^{n-r}.$$

\end{enumerate}
\end{proof}

Exercise 2.1.3

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Exercise 2.1.3

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