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

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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?

$已知数列{an}满足a1=1,(2n+5)(an+1)-(2n+7)an=4n^2+24n+35（n∈N+),则数列an$

(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)=1
a(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.
an/(2n+5)=1/7+1×(n-1)=(7n- 6)/7
an=(2n+5)(7n-6)/7
数列{an}的通项公式为an=(2n+5)(7n-6)/7.

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