数学 - 「離散的」な世界 - 数列 – 数学的帰納法と数列 - 帰納的定義(一般項)

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32、33.を取り組んでみる。

問32.

1. 
   $\begin{array}{l}{b}_n=a_{n+1}-a_n\\ =a_n+2n-a_n\\ =2n\\ \\ n>1\\ a_n=a_1+{\displaystyle \sum _{i=1}^{n-1}{b}_i}\\ =1+2·\frac{\left(n-1\right)n}{2}\\ =n^2-n+1\\ 1^2-1+1=1\\ \\ a_n=n^2-n+1\end{array}$
2. 
   $\begin{array}{l}{b}_n=a_{n+1}-a_n\\ =a_n+3^n-a_n\\ =3^n\\ \\ n>1\\ a_n=a_1+{\displaystyle \sum _{i=1}^{n-1}{b}_n}\\ =2+\frac{3\left(3^{n-1}-1\right)}{3-1}\\ =\frac{3^n}{2}+\frac{1}{2}\\ \\ \frac{3}{2}+\frac{1}{2}=2\\ a_n=\frac{1}{2}\left(3^n+1\right)\end{array}$
3. 
   $\begin{array}{l}x=3x-2\\ x=1\\ \\ a_{n+1}-1=3\left(a_n-1\right)\\ \\ a_n-1=\left(2-1\right)·3^{n-1}\\ a_n=3^{n-1}+1\end{array}$
4. 
   $\begin{array}{l}x=-2x+9\\ x=3\\ \\ a_{n+1}-3=-2\left(x-3\right)\\ \\ a_n-3=\left(1-3\right)·{\left(-2\right)}^{n-1}\\ a_n={\left(-2\right)}^n+3\end{array}$
5. 
   $\begin{array}{l}2x=x-6\\ x=-6\\ 2\left(a_{n+1}+6\right)=a_n+6\\ a_{n+1}+6=\frac{1}{2}\left(a_n+6\right)\\ \\ a_n+6=\left(2+6\right)·\frac{1}{2^{n-1}}\\ a_n=\frac{1}{2^{n-4}}-6\end{array}$
6. 
   $\begin{array}{l}1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5}\\ \\ a_{n+1}=\frac{n}{n+1}·\frac{1}{n}=\frac{1}{n+1}\\ \\ a_n=\frac{1}{n}\end{array}$
7. 
   $\begin{array}{l}2,\frac{2}{3},\frac{2}{5},\frac{2}{7},\frac{2}{9}\\ \\ a_{n+1}=\frac{\frac{2}{2n-1}}{\frac{2}{2n-1}+1}\\ =\frac{2}{2n-1}·\frac{2n-1}{2n+1`}\\ =\frac{2}{2\left(n+1\right)-1}\\ \\ a_n=\frac{2}{2n-1}\end{array}$
8. 
   $\begin{array}{l}3,3,6,12,24\\ \\ n>1\\ a_{n+1}=a_1+a_2+···+a_{n-1}+a_n\\ =2a_n\\ a_n=3·2^{n-2}\\ \\ a_1=3,a_n=3·2^{n-2}\left(n>1\right)\end{array}$

問33.

$\begin{array}{l}1,\frac{1}{3},\frac{1}{7},\frac{1}{15},\frac{1}{31}\\ \\ a_{n+1}=\frac{\frac{1}{2^n-1}}{\frac{1}{2^n-1}+2}\\ =\frac{1}{2^n-1}·\frac{2^n-1}{1+2^{n+1}-2}\\ =\frac{1}{2^{n+1}-1}\\ \\ a_n=\frac{1}{2^n-1}\end{array}$