已知数列{an}满足a1=1,(2n+5)(an+1)-(2n+7)an=4n^2+24n+35（n∈N+),则数列an的通项公式_数学解答

已知数列{an}满足a1=1,(2n+5)(an+1)-(2n+7)an=4n^2+24n+35（n∈N+),则数列an的通项公式为?

人气：414 ℃　时间：2020-02-04 01:37:23

解答

(2n+5)a(n+1)-(2n+7)an=4n²+24n+35=(2n+5)(2n+7)等式两边同除以(2n+5)(2n+7)a(n+1)/(2n+7)-an/(2n+5)=1a(n+1)/[2(n+1)+5]-an/(2n+5)=1,为定值.a1/(2×1+5)=1/7数列{an/(2n+5)}是以1/7为首项,1为公差的等差数列....