# 等比数列の和

• $r\ne 1$のとき

${S}_{n}=\sum _{m=1}^{n}{a}_{m}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}$$=\frac{{a}_{1}\left({r}^{n}-1\right)}{r-1}$

• $r=1$のとき

${S}_{n}=\sum _{m=1}^{n}{a}_{m}={a}_{1}n$

となる．

## ■公式の導出

$n$ 項までの和は

$\left[1\right]\text{\hspace{0.17em}}\text{\hspace{0.17em}}r=1$ のとき

${S}_{n}={a}_{1}+{a}_{1}+{a}_{1}+\cdot \cdot \cdot \cdot \cdot \cdot +{a}_{1}$

$=n{a}_{1}$

$\left[2\right]\text{\hspace{0.17em}}\text{\hspace{0.17em}}r\ne 1$ のとき

${S}_{n}={a}_{1}+{a}_{1}r+{a}_{1}{r}^{2}+{a}_{1}{r}^{3}+\cdot \cdot \cdot \cdot \cdot \cdot +{a}_{1}{r}^{n-1}$

$-\right)r{S}_{n}={a}_{1}r+{a}_{1}{r}^{2}+{a}_{1}{r}^{3}+\cdot \cdot \cdot \cdot \cdot \cdot +{a}_{1}{r}^{n-1}+{a}_{1}{r}^{n}$

$\overline{\text{\hspace{0.17em}\hspace{0.17em}}\left(1-r\right){S}_{n}={a}_{1}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-{a}_{1}{r}^{n}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}$

$\left(1-r\right){S}_{n}={a}_{1}\left(1-{r}^{n}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}$

${S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}=\frac{{a}_{1}\left({r}^{n}-1\right)}{r-1}$

よって，$\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[1\right]\text{\hspace{0.17em}}\text{\hspace{0.17em}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[2\right]\text{\hspace{0.17em}}\text{\hspace{0.17em}}$ より，第$n$ 項までの和は

• $r\ne 1$のとき

${S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}=\frac{{a}_{1}\left({r}^{n}-1\right)}{r-1}$

• $r=1$のとき

${S}_{n}={a}_{1}n$

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