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等比数列前n项求和公式证明

作者:互联网

\[设等比数列a_{n}=ar^{n-1},首项为a_{1},r为公比,n\in N^{*}.\\ 求其前n项之和(设为s_{n}) \]

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\[s_{n}=a_{1}+a_{2}+a_{3}+...+a_{n}=a_{1}r^0+a_{1}r^{1}+a_{1}r^{2}+...+a_{1}r^{n-1} \]

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\[设s_{u}=r \cdot s_{n} \\ =r(a_{1}r^{0}+a_{1}r^{1}+a_{1}r^{2}+...+a_{1}r^{n-1}) \\ =a_{1}r^{1}+a_{1}r^{2}+a_{1}r^{3}+...+a_{1}r^{n} \]

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\[比较s_{n}和s_{u}:\quad 发现若s_{n}移走a_{1}r^{0}, \quad s_{u}移走a_{1}r^{n}之后,\quad 余下部分重合 \]

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\[则s_{n}-s_{u}=s_{n}-s_{n}r \\=a_{1}-a_{1}r^{n} \\ =s_{n}(1-r) \\ =a_{1}(1-r^{n}) \]

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\[s_{n}=\frac{a_{1}(1-r^{n})}{1-r},r \ne 1 \]

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\[若r等于1,\\ s_{n}=a_{1}\cdot1^{0}+a_{1}\cdot1^{1}+a_{1}\cdot1^{2}+...+a_{1} \cdot 1^{n-1} =n \cdot a_{1} \]

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\[得出s_{n}的公式:\\ \]

\[当r=1时,s_{n}=a_{1}n \]

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\[当r\ne 1时,\quad s_{n}=\frac{a_{1}(1-r^{n})}{1-r} \]

标签:等比数列,求和,公式,frac,ne,cdot1,cdot,quad,+...+
来源: https://www.cnblogs.com/Preparing/p/16581231.html