等比数列前n项求和公式证明
作者：互联网
\[设等比数列a_{n}=ar^{n-1},首项为a_{1},r为公比,n\in N^{*}.\\
求其前n项之和(设为s_{n})
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\]\[s_{n}=a_{1}+a_{2}+a_{3}+...+a_{n}=a_{1}r^0+a_{1}r^{1}+a_{1}r^{2}+...+a_{1}r^{n-1}
\]\[\\ \\
\]\[设s_{u}=r \cdot s_{n}
\\
=r(a_{1}r^{0}+a_{1}r^{1}+a_{1}r^{2}+...+a_{1}r^{n-1})
\\
=a_{1}r^{1}+a_{1}r^{2}+a_{1}r^{3}+...+a_{1}r^{n}
\]\[\\ \\
\]\[比较s_{n}和s_{u}:\quad 发现若s_{n}移走a_{1}r^{0}, \quad s_{u}移走a_{1}r^{n}之后,\quad 余下部分重合
\]\[\\ \\
\]\[则s_{n}-s_{u}=s_{n}-s_{n}r
\\=a_{1}-a_{1}r^{n}
\\
=s_{n}(1-r)
\\
=a_{1}(1-r^{n})
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\]\[s_{n}=\frac{a_{1}(1-r^{n})}{1-r},r \ne 1
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\]\[若r等于1,\\
s_{n}=a_{1}\cdot1^{0}+a_{1}\cdot1^{1}+a_{1}\cdot1^{2}+...+a_{1}
\cdot 1^{n-1} =n \cdot a_{1}
\]\[\\ \\
\]\[得出s_{n}的公式:\\
\]\[当r=1时,s_{n}=a_{1}n
\]\[\\ \\
\]\[当r\ne 1时,\quad s_{n}=\frac{a_{1}(1-r^{n})}{1-r}
\]
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