展開･因数分解の公式(2変数編)

$$
\begin{array}{}
(ax+by)(cx+dy)&=&acx^2+(ad+bc)xy+bdy^2\\
(a+b)(a-b)&=&a^2-b^2\\
(ax+by)^n&=&\displaystyle\sum_{k=0}^{n}
\frac{n!}{k!(n-k)!}(ax)^k(by)^{n-k}\\
(a+b)\left[\displaystyle\sum_{k=0}^{n-1}a^k(-b)^{n-k-1}\right]&=&a^n-(-b)^n\cdots\cdots(\alpha)\\
(a-b)\left[\displaystyle\sum_{k=0}^{n-1}a^kb^{n-k-1}\right]&=&a^n-b^n\\
\end{array}
$$

$${(\alpha)}$$の証明

$$
\begin{array}{}
(a+b)\left[\displaystyle\sum_{k=0}^{n-1}a^k(-b)^{n-k-1}\right]&=&
\displaystyle\sum_{k=0}^{n-1}a^{k+1}(-b)^{n-k-1}
-\displaystyle\sum_{k=0}^{n-1}a^k(-b)^{n-k}\\
&=&a^n
+\displaystyle\sum_{k=0}^{n-2}a^{k+1}(-b)^{n-k-1}
-(-b)^n
-\displaystyle\sum_{k=1}^{n-1}a^k(-b)^{n-k}\\
&=&a^n-(-b)^n
+\displaystyle\sum_{k=1}^{n-1}a^{k}(-b)^{n-k}
-\displaystyle\sum_{k=1}^{n-1}a^k(-b)^{n-k}\\
&=&a^n-(-b)^n
\end{array}
$$

5番目の式は$${(\alpha)}$$において$${b\rightarrow{}-b}$$とすればよい