高考数学之数列必背技巧
\[
\frac{1}{a_{k}a_{k+1}}=\frac{1}{d}\left(\frac{1}{a_{k}}-\frac{1}{a_{k+1}} \right)
\]
\[ \frac{1}{a_{k}a_{k+m}}=\frac{1}{md}\left(\frac{1}{a_{k}}-\frac{1}{a_{k+m}} \right)
\]
\[
\frac{1}{\sqrt{a_{k}}+\sqrt{a_{k+1}}} =\frac{1}{d}(\sqrt{a_{k+1}}-\sqrt{a_{k}})
\]
\[
\frac{1}{\sqrt{a_{k}}+\sqrt{a_{k+m}}} =\frac{1}{md}(\sqrt{a_{k+m}}-\sqrt{a_{k}})
\]
\[
\frac{n+1}{n^{2}(n+2)^{2}} =\frac{1}{4}\left[ \frac{1}{n^{2}}-\frac{1}{(n+2)^{2}} \right]
\]
\[
\frac{2^{n}}{(2^{n}-1)(2^{n+1}-1)} =\left[ \frac{1}{(2^{n}-1)}-\frac{1}{(2^{n+1}-1)} \right]
\]
\[
\frac{n2^{n}}{(n+1)(n+2)} =\left[ \frac{2^{n+1}}{(n+2)}-\frac{2^{n}}{n+1)} \right]
\]
\]
posted on 2020-07-01 08:48 Indian_Mysore 阅读(70) 评论(0) 收藏 举报
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