已知数列an满足a1=0.5,an=a(n-1)+1/(n^2-1),则数列an的通项公式为？

来源：百度知道编辑：UC知道时间：2024/05/31 14:42:57

过程，答案

a(2n)=a(2n-1)+1/[(2n)^2-1]=a(2n-1)+1/[(2n+1)(2n-1)]=a(2n-1)+(1/2)[1/(2n-1)-1/(2n+1)]
a(2n+1)=a(2n)+1/[(2n+1)^2-1]=a(2n)+1/[(2n+2)(2n)]=a(2n)+(1/4)[1/(n)-1/(n+1)]=a(2n-1)+(1/2)[1/(2n-1)-1/(2n+1)]+(1/4)[1/n-1/(n+1)],
a(2n+1)+(1/2)[1/(2n+1)]+(1/4)[1/(n+1)]=a(2n-1)+(1/2)[1/(2n-1)]+(1/4)[1/n]=...=a(1)+(1/2)[1/1]+(1/4)[1/1]=1/2+1/2+1/4=5/4,
a(2n-1)=5/4-1/[2(2n-1)]-1/(4n),n=1,2,...
a(2n)=a(2n-1)+1/[2(2n-1)]-1/[2(2n+1)]=5/4-1/[2(2n-1)]-1/(4n)+1/[2(2n-1)]-1/[2(2n+1)]=5/4-1/[2(2n+1)]-1/(4n),n=1,2,...
综合，有
a(n)=5/4-1/(2n+2)-1/(2n),n=1,2,...

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