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战巡
Post time 2016-6-19 14:56
回复 1# lrh2006
这个不难吧
首先易证
\[a_n=\frac{2^{n+1}}{n^2+1}\]
\[c_n=\frac{1}{n(n+1)a_{n+1}}=\frac{1}{2^{n+2}}(1+\frac{2}{n}-\frac{1}{n+1})\]
\[S_n=\sum_{k=1}^nc_k=\sum_{k=1}^n\frac{1}{2^{k+2}}+\sum_{k=1}^n(\frac{1}{2^{k+1}k}-\frac{1}{2^{k+2}(k+1)})=\frac{1}{4}-\frac{1}{2^{n+2}}+\frac{1}{4}-\frac{1}{2^{n+2}(n+1)}\]
\[=\frac{1}{2}-\frac{1}{2^{n+2}}-\frac{1}{2^{n+2}(n+1)}\] |
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