## 2017年1月12日木曜日

### 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

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

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

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

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