成都市14级“二诊”16题的解法探究

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在数列({a_{_n}})中，(a_{_1}=1),(a_{_n}=dfrac{n^2}{n^2-1}a_{_{n-1}}(ngeqslant 2,nin N^*)),则数列({dfrac{a_{_{n}}}{n^2}})的前项和(T_{_n}=underline{qquadlacktriangleqquad}.)

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我的解法：

(a_{_n}=dfrac{n^2}{n^2-1}a_{_{n-1}}Rightarrow n^2a_n-a_n=n^2a_{n-1}Rightarrow dfrac{a_n}{n^2}=a_n-a_{n-1})

(Rightarrow dfrac{a_1}{1^2}=a_1=1,dfrac{a_2}{2^2}=a_2-a_{1},dfrac{a_3}{3^2}=a_3-a_{2},cdots,dfrac{a_n}{n^2}=a_n-a_{n-1}Rightarrow T_n=a_n)

(a_1=1,a_{_n}=dfrac{n^2}{n^2-1}a_{_{n-1}}Rightarrow a_1=1,a_2=dfrac{4}{3},a_3=dfrac{6}{4},a_4=dfrac{8}{5})

归纳出(a_n=dfrac{2n}{n+1})

当然，也可以先归纳出(a_n=dfrac{2n}{n+1})(低成本探究“列举法”)，再求(dfrac{a_n}{n^2}).

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南部中学吴老师的解法：

令(b_n=dfrac{a_n}{n^2}), (a_1=1,a_{_n}=dfrac{n^2}{n^2-1}a_{n-1}Rightarrow b_n=b_{n-1}dfrac{n-1}{n+1}(ngeqslant 2))

(Rightarrow dfrac{b_n}{b_{n-1}}=dfrac{n-1}{n+1}Rightarrow b_n=dfrac{2}{(n+1)n}=dfrac{2}{n}-dfrac{2}{n+1}Rightarrow T_n=b_1+b_2+cdots+b_n=dfrac{2n}{n+1})

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达州周老师的解法：

(a_n=dfrac{nn}{(n-1)(n+1)}a_{n-1}Rightarrow dfrac{n+1}{n}a_n=dfrac{n}{n-1}a_{n-1}=cdots=dfrac{3}{2}a_2=dfrac{2}{1}a_1=2)

(Rightarrow a_n=dfrac{2n}{n+1})

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乐山陈朝斌老师的解法：

(dfrac{a_n}{a_{n-1}}=dfrac{nn}{(n-1)(n+1)})

(Rightarrow dfrac{a_2}{a_{1}} imesdfrac{a_3}{a_{2}} imesdfrac{a_4}{a_{3}} imescdots imesdfrac{a_n}{a_{n-1}}=(dfrac{2}{3}cdotdfrac{3}{4}cdotdfrac{4}{5}cdotsdfrac{n}{n+1})(dfrac{2}{1}cdotdfrac{3}{2}cdotdfrac{4}{3}cdotsdfrac{n}{n-1})=dfrac{2n}{n+1})

(Rightarrow a_n=dfrac{2n}{n+1})