数列 $\{a_{n}\}$ 满足：$a_{1}=1$，$\displaystyle a_{n+1}=1+\dfrac{1}{n}\sum\limits_{k=1}^{n}a_{k}$．
【难度】
【出处】
2013年全国高中数学联赛山西省预赛
【标注】
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写出数列前 $7$ 项的值；标注答案$a_{1}=1$，$a_{2}=2$，$a_{3}=\dfrac{5}{2}$，$a_{4}=\dfrac{17}{6}$，$a_{5}=\dfrac{37}{12}$，$a_{6}=\dfrac{197}{60}$，$a_{7}=\dfrac{207}{60}$解析略
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对任意正整数 $n$，求 $a_{n}$ 的表达式．标注答案$a_{n}=1+\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n-1}\right)$解析因为\[\begin{split}a_{n+1}&=1+\dfrac{a_{1}+a_{2}+\cdots+a_{n-1}+a_{n}}{n},\\ a_{n}&=1+\dfrac{a_{1}+a_{2}+\cdots+a_{n-1}}{n-1},\end{split}\]所以\[\begin{split}a_{n+1}-a_{n}&=\dfrac{a_{1}+a_{2}+\cdots+a_{n-1}+a_{n}}{n}-\dfrac{a_{1}+a_{2}+\cdots+a_{n-1}}{n-1}\\&=\dfrac{a_{n}}{n}+\left(\dfrac{a_{1}+a_{2}+\cdots+a_{n-1}}{n}-\dfrac{a_{1}+a_{2}+\cdots+a_{n-1}}{n-1}\right)\\&=\dfrac{1}{n}\left(1+\dfrac{a_{1}+a_{2}+\cdots+a_{n-1}}{n-1}\right)+(a_{1}+a_{2}+\cdots+a_{n-1})\left(\dfrac{1}{n}-\dfrac{1}{n-1}\right)\\&=\dfrac{1}{n}.\end{split}\]由\[\begin{split}a_{n}-a_{n-1}&=\dfrac{1}{n-1},\\ a_{n-1}-a_{n-2}&=\dfrac{1}{n-2},\\ \cdots&\cdots\\ a_{2}-a_{1}&=\dfrac{1}{1},\\ a_{1}&=1,\end{split}\]相加得\[a_{n}=1+\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n-1}\right).\]
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