二項定理

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ただいま作成中

私の授業で使いながら問題を増やしているため、完成するまでに時間がかかりそうです。少しずつ問題を増やしたり、ポイント解説を付けたりしていきます。無限の彼方で完成する日を、どうぞご期待ください。

Happy Math-ing!

未完成でもよければ、使ってやってください。😃

次の式を展開しよう。

↓この問題へのリンクはこちら（右クリックで保存）

【解答】二項定理を使って解く

\def\SL{x}
\def\SF{}
\def\SR{-2}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& (\colBX{bisque}{$\SL$}\SF\colBX{palegreen}{$\SR$})^{\colBX{mistyrose}{$\scriptsize 5$}}\\
\\
&= {}_{5}{\rm C}_{0}(\colBX{bisque}{$\SL$})^{\colBX{mistyrose}{$\scriptsize 5$}}+{}_{5}{\rm C}_{1}(\colBX{bisque}{$\SL$})^4(\colBX{palegreen}{$\SR$})^{1}+{}_{5}{\rm C}_{2}(\colBX{bisque}{$\SL$})^3(\colBX{palegreen}{$\SR$})^{2}+{}_{5}{\rm C}_{3}(\colBX{bisque}{$\SL$})^2(\colBX{palegreen}{$\SR$})^{3}+{}_{5}{\rm C}_{4}(\colBX{bisque}{$\SL$})^1(\colBX{palegreen}{$\SR$})^{4}+{}_{5}{\rm C}_{5}(\colBX{palegreen}{$\SR$})^{\colBX{mistyrose}{$\scriptsize 5$}}\\
\\
&= 1x^5+5x^4 \cdot (-2)+10x^3 \cdot 4 +10x^2 \cdot (-8) + 5 x \cdot 16 + 1 \cdot (-32)\\
\\
&= x^5-10x^4+40x^3-80x^2+80x-32
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】筆算で解く

\def\SL{x}
\def\SF{}
\def\SR{-2}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{matrix}
(\colBX{bisque}{$\SL$}\SF\colBX{palegreen}{$\SR$})^{\colBX{mistyrose}{$\scriptsize 5$}}\\
\\
& (\colBX{bisque}{$\SL$})^{\colBX{mistyrose}{$\scriptsize 5$}} & (\colBX{bisque}{$\SL$})^{4} & (\colBX{bisque}{$\SL$})^{3} & (\colBX{bisque}{$\SL$})^{2} & (\colBX{bisque}{$\SL$})^{1} & \\
& & (\colBX{palegreen}{$\SR$})^1 &  (\colBX{palegreen}{$\SR$})^2 &  (\colBX{palegreen}{$\SR$})^3 &  (\colBX{palegreen}{$\SR$})^4 &  (\colBX{palegreen}{$\SR$})^{\colBX{mistyrose}{$\scriptsize 5$}}\\
\times ) & 1 & 5 & 10 & 10 & 5  & 1\\\hline
& x^5 & -10x^4 & +40x^3 & -80x^2 & +80x & -32 
\end{matrix}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】頑張って展開して解く

\def\SL{x}
\def\SF{}
\def\SR{-2}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& (\SL\SF\SR)^5\\
\\
&= (x-2)^3(x-2)^2\\
\\
&= (x^3-6x^2+12x-8)(x^2-4x+4)\\
\\
&= (x^3-6x^2+12x-8)x^2+(x^3-6x^2+12x-8)(-4x)+(x^3-6x^2+12x-8) \cdot 4\\
\\
&= (x^5-6x^4+12x^3-8x^2)+(-4x^4+24x^3-48x^2+32x)+(4x^3-24x^2+48x-32)\\
\\
&= x^5-10x^4+40x^3-80x^2+80x-32
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

↓この問題へのリンクはこちら（右クリックで保存）

【解答】二項定理を使って解く

\def\SL{x}
\def\SF{+}
\def\SR{1}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& (\colBX{bisque}{$\SL$}\SF\colBX{palegreen}{$\SR$})^{\colBX{mistyrose}{$\scriptsize 4$}}\\
\\
&= {}_{4}{\rm C}_{0}(\colBX{bisque}{$\SL$})^{\colBX{mistyrose}{$\scriptsize 4$}}+{}_{4}{\rm C}_{1}(\colBX{bisque}{$\SL$})^3(\colBX{palegreen}{$\SR$})^{1}+{}_{4}{\rm C}_{2}(\colBX{bisque}{$\SL$})^2(\colBX{palegreen}{$\SR$})^{2}+{}_{4}{\rm C}_{3}(\colBX{bisque}{$\SL$})^1(\colBX{palegreen}{$\SR$})^{3}+{}_{4}{\rm C}_{4}(\colBX{palegreen}{$\SR$})^{\colBX{mistyrose}{$\scriptsize 4$}}\\
\\
&= 1x^4+4x^3 \cdot 1+6x^2 \cdot 1 +4x^1 \cdot 1 + 1 \cdot 1\\
\\
&= x^4+4x^3+6x^2+4x+1
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】筆算で解く

\def\SL{x}
\def\SF{+}
\def\SR{1}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{matrix}
(\colBX{bisque}{$\SL$}\SF\colBX{palegreen}{$\SR$})^{\colBX{mistyrose}{$\scriptsize 4$}}\\
\\
& (\colBX{bisque}{$\SL$})^{\colBX{mistyrose}{$\scriptsize 4$}} & (\colBX{bisque}{$\SL$})^{3} & (\colBX{bisque}{$\SL$})^{2} & (\colBX{bisque}{$\SL$})^{1} & \\
& & (\colBX{palegreen}{$\SR$})^1 &  (\colBX{palegreen}{$\SR$})^2 &  (\colBX{palegreen}{$\SR$})^3 &  (\colBX{palegreen}{$\SR$})^4\\
\times ) & 1 & 4 & 6 & 4 & 1\\\hline
& x^4 & +4x^3 & +6x^2 & +4x & +1 
\end{matrix}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】頑張って展開して解く

\def\SL{x}
\def\SF{+}
\def\SR{1}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& (\SL\SF\SR)^4\\
\\
&= (x+1)^2(x+1)^2\\
\\
&= (x^2+2x+1)(x^2+2x+1)\\
\\
&= (x^2+2x+1)x^2+(x^2+2x+1)(2x)+(x^2+2x+1) \cdot 1\\
\\
&= (x^4+2x^3+x^2)+(2x^3+4x^2+2x)+(x^2+2x+1)\\
\\
&= x^4+4x^3+6x^2+4x+1
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

↓この問題へのリンクはこちら（右クリックで保存）

【解答】二項定理を使って解く

\def\SL{x}
\def\SF{}
\def\SR{-2}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& (\colBX{bisque}{$\SL$}\SF\colBX{palegreen}{$\SR$})^{\colBX{mistyrose}{$\scriptsize 6$}}\\
\\
&= {}_{6}{\rm C}_{0}(\colBX{bisque}{$\SL$})^{\colBX{mistyrose}{$\scriptsize 6$}}+{}_{6}{\rm C}_{1}(\colBX{bisque}{$\SL$})^5(\colBX{palegreen}{$\SR$})^{1}+{}_{6}{\rm C}_{2}(\colBX{bisque}{$\SL$})^4(\colBX{palegreen}{$\SR$})^{2}+{}_{6}{\rm C}_{3}(\colBX{bisque}{$\SL$})^3(\colBX{palegreen}{$\SR$})^{3}+{}_{6}{\rm C}_{4}(\colBX{bisque}{$\SL$})^2(\colBX{palegreen}{$\SR$})^{4}+{}_{6}{\rm C}_{5}(\colBX{bisque}{$\SL$})^1(\colBX{palegreen}{$\SR$})^{\colBX{mistyrose}{$\scriptsize 5$}}+{}_{6}{\rm C}_{6}(\colBX{palegreen}{$\SR$})^{\colBX{mistyrose}{$\scriptsize 5$}}\\
\\
&= 1x^6+6x^5 \cdot (-2)+15x^4 \cdot 4 +20x^3 \cdot (-8) + 15 x^2 \cdot 16 + 6x \cdot (-32)+1 \cdot 64\\
\\
&= x^6-12x^5+60x^4-160x^3+240x^2-192x+64
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】筆算で解く

\def\SL{x}
\def\SF{}
\def\SR{-2}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{matrix}
(\colBX{bisque}{$\SL$}\SF\colBX{palegreen}{$\SR$})^{\colBX{mistyrose}{$\scriptsize 6$}}\\
\\
& (\colBX{bisque}{$\SL$})^{\colBX{mistyrose}{$\scriptsize 6$}} & (\colBX{bisque}{$\SL$})^{5} & (\colBX{bisque}{$\SL$})^{4} & (\colBX{bisque}{$\SL$})^{3} & (\colBX{bisque}{$\SL$})^{2} & (\colBX{bisque}{$\SL$})^{1} & \\
& & (\colBX{palegreen}{$\SR$})^1 &  (\colBX{palegreen}{$\SR$})^2 &  (\colBX{palegreen}{$\SR$})^3 &  (\colBX{palegreen}{$\SR$})^4 &  (\colBX{palegreen}{$\SR$})^5 &  (\colBX{palegreen}{$\SR$})^{\colBX{mistyrose}{$\scriptsize 6$}}\\
\times ) & 1 & 6 & 15 & 20 & 15  & 6 & 1\\\hline
& x^6 & -12x^5 & +60x^4 & -160x^3 & +240x^2 & -192x & +64 
\end{matrix}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】頑張って展開して解く

\def\SL{x}
\def\SF{}
\def\SR{-2}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& (\SL\SF\SR)^6\\
\\
&= (x-2)^3(x-2)^3\\
\\
&= (x^3-6x^2+12x-8)(x^3-6x^2+12x-8)\\
\\
&= (x^3-6x^2+12x-8)x^3+(x^3-6x^2+12x-8)(-6x^2)+(x^3-6x^2+12x-8) \cdot 12x+(x^3-6x^2+12x-8) \cdot (-8)\\
\\
&= (x^6-6x^5+12x^4-8x^3)+(-6x^5+36x^4-72x^3+48x^2)+(12x^4-72x^3+144x^2-96x)+(-8x^3+48x^2-96x+64)\\
\\
&= x^6-12x^5+60x^4-160x^3+240x^2-192x+64
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

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