等差数列の和

# 等差数列の和

${S}_{n}=\sum _{m=1}^{n}{a}_{m}=\frac{n\left\{2{a}_{1}+\left(n-1\right)d\right\}}{2}$$=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}$

となる．

## ■公式の導出

$n$ 項までの和は

${S}_{n}={a}_{1}+{a}_{2}+{a}_{3}+\cdots \cdots +{a}_{n-1}+{a}_{n}$

$={a}_{1}+\left({a}_{1}+d\right)+\left({a}_{1}+2d\right)+\cdot \cdot \cdot \cdot \cdot \cdot +\left({a}_{n}-d\right)+{a}_{n}$

となる．

${S}_{n}={a}_{1}+\left({a}_{1}+d\right)+\left({a}_{1}+2d\right)+\cdot \cdot \cdot \cdot \cdot \cdot +\left({a}_{n}-d\right)+{a}_{n}$

$+\right)\text{\hspace{0.17em}}{S}_{n}={a}_{n}+\left({a}_{n}-d\right)+\left({a}_{n}-2d\right)+\cdot \cdot \cdot \cdot \cdot \cdot +\left({a}_{1}+d\right)+{a}_{1}$

$\overline{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2{S}_{n}=\left({a}_{1}+{a}_{n}\right)+\left({a}_{1}+{a}_{n}\right)+\left({a}_{1}+{a}_{n}\right)+\cdot \cdot \cdot +\left({a}_{1}+{a}_{n}\right)+\left({a}_{1}+{a}_{n}\right)\text{\hspace{0.17em}}}$

$2{S}_{n}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=n\left({a}_{1}+{a}_{n}\right)$

${S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}$

また，$\text{\hspace{0.17em}}\text{\hspace{0.17em}}{a}_{n}={a}_{1}+\left(n-1\right)d\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}$ を代入すると

${S}_{n}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{n\left\{{a}_{1}+{a}_{1}+\left(n-1\right)d\right\}}{2}$ $=\frac{n\left\{2{a}_{1}+\left(n-1\right)d\right\}}{2}$

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