Question

已知正项数列[an}满足:a1=3,(2n-1)an+2=(2n+1)an-1+8n^2（n＞1,n∈N*）求数列{an？

Answer 1

已知正项数列{an},满足a1=3,(2n-1)an+2=(2n+1)an-1+8n^2（n>1n属于正数）1、求an通项公式 2、设bn=1/an,求数列bn前n项和
1.(2n-1)an+2=(2n+1)an-1+8n^2
an=(2n+1)/(2n-1)*an-1+(8n^2-2)/(2n-1)
即an=(2n+1)/(2n-1)*an-1+4n+2
两边同时除以2n+1得
an/(2n+1)=an-1/(2n-1)+2
所以数列an/(2n+1)是等差数列
首项a1/(2+1)=1,公差为2
所以an/(2n+1)=2n-1
an通项公式是an=(2n+1)(2n-1)=4n^2-1
(2)bn=1/an=1/((2n+1)(2n-1))=1/2(1/(2n-1)-1/(2n+1))
b1=1/2(1-1/3)
b2=1/2(1/3-1/5)
……
bn=1/2(1/(2n-1)-1/(2n+1))
累加得数列bn前n项和=1/2(1-1/(2n+1))=n/(2n+1),3,已知正项数列[an}满足:a1=3,(2n-1)an+2=(2n+1)an-1+8n^2（n＞1,n∈N*）求数列{an}的通项an
设bn=1/an,求数列{bn}的前n项的和Sn