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?
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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?](/uploads/image/z/12641804-44-4.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D1%2C%282n%2B5%29%28an%2B1%29-%282n%2B7%29an%3D4n%5E2%2B24n%2B35%EF%BC%88n%E2%88%88N%2B%29%2C%E5%88%99%E6%95%B0%E5%88%97an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E4%B8%BA%3F)
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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的通项公式为?
(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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