数学 - 「離散的」な世界 - 数列 – 数学的帰納法と数列 - 数学的帰納法

数学読本〈3〉

学習環境

• Surface 3 (4G LTE)、Surface 3 タイプ カバー、Surface ペン(端末)
• Windows 10 Pro (OS)
• 数式入力ソフト(TeX, MathML): MathType
• MathML対応ブラウザ: Firefox、Safari
• MathML非対応ブラウザ(Internet Explorer, Google Chrome...)用JavaScript Library: MathJax

数学読本〈3〉平面上のベクトル/複素数と数学的帰納法と数列/空間図形/2次曲線/数列(松坂 和夫(著)、岩波書店)の第13章(「離散的」な世界 - 数列)、13.2(数学的帰納法と数列)、数学的帰納法、問27、28、29.を取り組んでみる。

問27.

$\begin{array}{l}{a}^n<b^n\\ a^{n+1}<a{b}^n<b{b}^n=b^{n+1}\\ a^{n+1}<b^{n+1}\end{array}$

問28.

1. 
   $\begin{array}{l}n=1\\ \frac{1}{6}·2·3=1\\ \\ \frac{1}{6}n\left(n+1\right)\left(2n+1\right)+{\left(n+1\right)}^2\\ =\frac{\left(n+1\right)\left(n\left(2n+1\right)+6\left(n+1\right)\right)}{6}\\ =\frac{1}{6}\left(n+1\right)\left(2n^2+7n+6\right)\\ =\frac{1}{6}\left(n+1\right)\left(n+2\right)\left(2n+3\right)\end{array}$
2. 
   $\begin{array}{l}\frac{1}{1·2}=\frac{1}{1+1}\\ \\ \frac{1}{1·2}+···+\frac{1}{n\left(n+1\right)}+\frac{1}{\left(n+1\right)\left(n+2\right)}\\ =\frac{n}{n+1}+\frac{1}{\left(n+1\right)\left(n+2\right)}\\ =\frac{n\left(n+2\right)+1}{\left(n+1\right)\left(n+2\right)}\\ =\frac{n^2+2n+1}{\left(n+1\right)\left(n+2\right)}\\ =\frac{{\left(n+1\right)}^2}{\left(n+1\right)\left(n+2\right)}\\ =\frac{n+1}{n+2}\end{array}$
3. 
   $\begin{array}{l}1·2·3=6\\ \frac{1}{4}·1·2·3·4=6\\ \\ 1·2·3+···+n\left(n+1\right)\left(n+2\right)+\left(n+1\right)\left(n+2\right)\left(n+3\right)\\ =\frac{1}{4}n\left(n+1\right)\left(n+2\right)\left(n+3\right)+\left(n+1\right)\left(n+2\right)\left(n+3\right)\\ =\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}{4}\end{array}$
4. 
   $\begin{array}{l}1-\frac{1}{2}=\frac{1}{2}\\ \\ 1-\frac{1}{2}+···+\frac{1}{2n-1}-\frac{1}{2n}+\frac{1}{2n+1}-\frac{1}{2n+2}\\ =\frac{1}{n+1}+\frac{1}{n+2}+···+\frac{1}{2n}+\frac{1}{2n+1}-\frac{1}{2n+2}\\ =\frac{1}{n+2}+···+\frac{1}{2n}+\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{n+1}\\ =\frac{1}{n+2}+···+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}\end{array}$
5. 
   $\begin{array}{l}{1}^2-2^2=-3\\ -1·3=-3\\ \\ 1^2-2^2+···+{\left(2n-1\right)}^2-{\left(2n\right)}^2+{\left(2n+1\right)}^2-{\left(2n+2\right)}^2\\ =-n\left(2n+1\right)+{\left(2n+1\right)}^2-{\left(2n+2\right)}^2\\ =-2n^2-5n-3\\ =-\left(n+1\right)\left(2n+3\right)\end{array}$

問29.

$\begin{array}{l}8-7-1=0\\ \\ 8^{n+1}-7\left(n+1\right)-1\\ =8·8^n-7n-7-1\\ =8·8^n-7n-8\\ =8\left(8^n-7n-1\right)+49n\end{array}$