S-rSを書き並べる必要はない（その2）

$$f(n)=a\{(n+1)-n\}$$
よって$g(n)=an$のとき$f(n)=g(n+1)-g(n)$

$$\begin{array}{rcl}f(n)& =& \frac{a}{2}(2n+1)+(b-\frac{a}{2})\\ & =& \frac{a}{2}\{(n+1{)}^2-n^2\}+(b-\frac{a}{2})\{(n+1)-n\}\end{array}$$
$$\therefore g(n)=\frac{a}{2}{n}^2+(b-\frac{a}{2})n$$
$(n+1{)}^2-n^2=2n+1$となることを意識して変形すると良い。

$$\begin{array}{rcl}f(n)& =& \frac{r-1}{r-1}\cdot r^n\\ & =& \frac{r^{n+1}-r^n}{r-1}\\ & =& \frac{r^{n+1}}{r-1}-\frac{r^n}{r-1}\end{array}$$
$\begin{array}{rcl}\therefore g(n)& =& \frac{r^n}{r-1}\\ & =& \frac{r}{r-1}\cdot r^{n-1}\end{array}$

$$\begin{array}{rcl}f(n)& =& \frac{r-1}{r-1}\cdot n{r}^n\\ & =& \frac{n\cdot r^{n+1}}{r-1}-\frac{n\cdot r^n}{r-1}\\ & =& \frac{\{(n+1)-1\}\cdot r^{n+1}}{r-1}-\frac{n\cdot r^n}{r-1}\\ & =& \frac{(n+1)\cdot r^{n+1}}{r-1}-\frac{n\cdot r^n}{r-1}-\frac{r^{n+1}}{r-1}\\ & =& \frac{(n+1)\cdot r^{n+1}}{r-1}-\frac{n\cdot r^n}{r-1}-\frac{1}{r-1}(\frac{r^{n+2}}{r-1}-\frac{r^{n+1}}{r-1})\end{array}$$
$$\begin{array}{rcl}\therefore g(n)& =& \frac{n\cdot r^n}{r-1}-\frac{1}{r-1}\left(\frac{r^{n+1}}{r-1}\right)\\ & =& \frac{r}{r-1}(n-{\displaystyle \frac{r}{r-1}})\cdot r^{n-1}\end{array}$$