成都市14级“二诊”16题的解法探究 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}.) 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}). 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}) 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}) 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}) 每周看看我,冲进985!【魏刚的作品,转载须声明】