## 2016年12月29日木曜日

### 数学 - 「離散的」な世界 - 数列 – 数列とその和 - 平方の和、立方の和(一般項)

1. $3\frac{n\left(n+1\right)}{2}-2n=\frac{n\left(3n-1\right)}{2}$

2. $\begin{array}{l}\sum _{i=1}^{n}\left(2{i}^{2}+i-1\right)\\ =\frac{2n\left(2n+1\right)\left(n+1\right)}{6}+\frac{n\left(n+1\right)}{2}-n\\ =\frac{n\left(4{n}^{2}+6n+2+3n+3-6\right)}{6}\\ =\frac{n\left(4{n}^{2}+9n-1\right)}{6}\end{array}$

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

4. $\begin{array}{l}\frac{n\left(2n+1\right)\left(n+1\right)}{6}+\frac{3n\left(n+1\right)}{2}-4n\\ =\frac{n\left(2{n}^{2}+3n+1+9n+9-24\right)}{6}\\ =\frac{n\left(2{n}^{2}+12n-14\right)}{6}\\ =\frac{n\left({n}^{2}+6n-7\right)}{3}\\ =\frac{n\left(n+7\right)\left(n-1\right)}{3}\end{array}$
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6. $\begin{array}{l}{\left(\frac{n\left(n+1\right)}{2}\right)}^{2}+\frac{2n\left(2n+1\right)\left(n+1\right)}{6}+n\\ =\frac{{n}^{2}{\left(n+1\right)}^{2}}{4}+\frac{2n\left(2n+1\right)\left(n+1\right)}{6}+\frac{n\left(n+1\right)}{2}\\ =\frac{n\left(n+1\right)\left(3n\left(n+1\right)+4\left(2n+1\right)+6\right)}{12}\\ =\frac{n\left(n+1\right)\left(3{n}^{2}+3n+8n+4+6\right)}{12}\\ =\frac{n\left(n+1\right)\left(3{n}^{2}+11n+10\right)}{12}\\ =\frac{n\left(n+1\right)\left(n+2\right)\left(3n+5\right)}{12}\end{array}$

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

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

2. $\begin{array}{l}{a}_{n}={\left(2n\right)}^{2}=4{n}^{2}\\ \frac{4n\left(2n+1\right)\left(n+1\right)}{6}=\frac{2n\left(2n+1\right)\left(n+1\right)}{3}\end{array}$

3. $\begin{array}{l}{a}_{n}={n}^{2}\left(3n-1\right)=3{n}^{3}-{n}^{2}\\ \frac{3{n}^{2}{\left(n+1\right)}^{2}}{4}-\frac{n\left(2n+1\right)\left(n+1\right)}{6}\\ =\frac{n\left(n+1\right)\left(9n\left(n+1\right)-2\left(2n+1\right)\right)}{12}\\ =\frac{n\left(n+1\right)\left(9{n}^{2}+9n-4n-2\right)}{12}\\ =\frac{n\left(n+1\right)\left(9{n}^{2}+5n-2\right)}{12}\end{array}$

4. $\begin{array}{l}{a}_{n}=\frac{n\left(n+1\right)}{2}=\frac{1}{2}\left({n}^{2}+n\right)\\ \frac{1}{2}\left(\frac{n\left(2n+1\right)\left(n+1\right)}{6}+\frac{n\left(n+1\right)}{2}\right)\\ =\frac{2n\left(n+1\right)\left(n+2\right)}{12}\\ =\frac{n\left(n+1\right)\left(n+2\right)}{6}\end{array}$

5. $\begin{array}{l}\frac{\frac{n\left(n+1\right)}{2}\left(\frac{n\left(n+1\right)}{2}+1\right)}{2}\\ =\frac{1}{8}n\left(n+1\right)\left(n\left(n+1\right)+2\right)\\ =\frac{n\left(n+1\right)\left({n}^{2}+n+2\right)}{8}\end{array}$