在数列\(\{a_{_n}\}\)中，\(a_{_1}=1\),\(a_{_n}=\dfrac{n^2}{n^2-1}a_{_{n-1}}(n\geqslant 2,n\in N^*)\),则数列\(\{\dfrac{a_{_{n}}}{n^2}\}\)的前项和\(T_{_n}=\underline{\qquad\blacktriangle\qquad}.\)
\(a_{_n}=\dfrac{n^2}{n^2-1}a_{_{n-1}}\Rightarrow n^2a_n-a_n=n^2a_{n-1}\Rightarrow \dfrac{a_n}{n^2}=a_n-a_{n-1}\)
\(\Rightarrow \dfrac{a_1}{1^2}=a_1=1,\dfrac{a_2}{2^2}=a_2-a_{1},\dfrac{a_3}{3^2}=a_3-a_{2},\cdots,\dfrac{a_n}{n^2}=a_n-a_{n-1}\Rightarrow T_n=a_n\)
\(a_1=1,a_{_n}=\dfrac{n^2}{n^2-1}a_{_{n-1}}\Rightarrow a_1=1,a_2=\dfrac{4}{3},a_3=\dfrac{6}{4},a_4=\dfrac{8}{5}\)
归纳出\(a_n=\dfrac{2n}{n+1}\)
令\(b_n=\dfrac{a_n}{n^2}\), \(a_1=1,a_{_n}=\dfrac{n^2}{n^2-1}a_{n-1}\Rightarrow b_n=b_{n-1}\dfrac{n-1}{n+1}(n\geqslant 2)\)
\(\Rightarrow \dfrac{b_n}{b_{n-1}}=\dfrac{n-1}{n+1}\Rightarrow b_n=\dfrac{2}{(n+1)n}=\dfrac{2}{n}-\dfrac{2}{n+1}\Rightarrow T_n=b_1+b_2+\cdots+b_n=\dfrac{2n}{n+1}\)
\(a_n=\dfrac{nn}{(n-1)(n+1)}a_{n-1}\Rightarrow \dfrac{n+1}{n}a_n=\dfrac{n}{n-1}a_{n-1}=\cdots=\dfrac{3}{2}a_2=\dfrac{2}{1}a_1=2\)
\(\Rightarrow a_n=\dfrac{2n}{n+1}\)
\(\dfrac{a_n}{a_{n-1}}=\dfrac{nn}{(n-1)(n+1)}\)
\(\Rightarrow \dfrac{a_2}{a_{1}}\times\dfrac{a_3}{a_{2}}\times\dfrac{a_4}{a_{3}}\times\cdots\times\dfrac{a_n}{a_{n-1}}=(\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdots\dfrac{n}{n+1})(\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdots\dfrac{n}{n-1})=\dfrac{2n}{n+1}\)
