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[数列] 求通项两题

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hongxian posted 2016-4-27 08:44 |Read mode
$ a_{n+1}=3^{a_n} $,$a_1=1$,求$a_n$



$a_{n+1}=na_n+1$,$a_1=1$,求$a_n$

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战巡 posted 2016-4-27 08:54
回复 1# hongxian


第一个没啥意思,直接迭代就完事了,那个式子我估计也没啥简化表达方式

第二个参见
blog.sina.com.cn/s/blog_7ac9421701016w89.html

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敬畏数学 posted 2016-4-27 20:41
这是哪里出来的啊?

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kuing posted 2016-4-27 21:01
要么是自己随便出的,要么是题目并没要求求通项的

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original poster hongxian posted 2016-4-27 21:46
回复 3# 敬畏数学

只是在一个群内看到的,感觉不会就发上来了!

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敬畏数学 posted 2016-4-27 21:50
[b]回复 4# kuing
有理。。

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abababa posted 2016-4-29 11:01
回复 1# hongxian
第二题等式就是$a_n=(n-1)a_{n-1}+1$,可以先解$a_{n}=(n-1)a_{n-1}$,解出$a_n=(n-1)!$,然后令原等式中的$a_n=(n-1)!b_n$,这样等式$a_n=(n-1)a_{n-1}+1$就是$(n-1)!b_n=(n-1)(n-2)!b_{n-1}+1$,就是$b_n-b_{n-1}=\frac{1}{(n-1)!}$,然后分别写出$n=1$到$n$的等式,相加就得$b_n-b_1=\sum_{i=1}^{n-1}\frac{1}{i!}$,这样$b_n=\sum_{i=1}^{n-1}\frac{1}{i!}+1$,再代回$a_n=(n-1)!b_n$,就得最后结果了。

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