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Author |
isee
Posted 2020-7-11 18:38
回复 5# 色k
每个人处理(2)中含根式方式均不同,我因为不愿意平方,这样写:
\begin{align*}
\sqrt{S_{n+1}}-\sqrt{S_n}&=\frac 1{\sqrt 3}\sqrt{a_{n+1}}\\[1em]
\sqrt{S_{n+1}}-\sqrt{S_n}&=\frac 1{\sqrt 3}\sqrt{S_{n+1}-S_n}\\[1em]
\frac{\sqrt{S_{n+1}}}{\sqrt{S_{n+1}-S_n}}-\frac{\sqrt{S_n}}{\sqrt{S_{n+1}-S_n}}&=\frac 1{\sqrt 3}\\[1em]
\frac{1}{\sqrt{1-X}}&=\frac{1}{\sqrt{\frac 1X-1}}+\frac 1{\sqrt 3},X=\frac {S_n}{S_{n+1}}\\[1em]
\frac{1}{1-X}&=\frac{1}{\frac 1X-1}+\frac 1{3}+\frac 2{\sqrt{3}}\cdot \frac{1}{\sqrt{\frac 1X-1}}\\[1em]
\frac{2}{3}&=\frac 2{\sqrt{3}}\cdot \frac{1}{\sqrt{\frac 1X-1}}\\[1em]
\frac 1X-1&=3
\end{align*}
下略 |
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