求数列的通项公式(累乘法)已知数列{an}中,a1=1,(2n+1)an=(2n-3)an-1(n≥2),则数列{an}的通项公式
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![求数列的通项公式(累乘法)已知数列{an}中,a1=1,(2n+1)an=(2n-3)an-1(n≥2),则数列{an}的通项公式](/uploads/image/z/667290-66-0.jpg?t=%E6%B1%82%E6%95%B0%E5%88%97%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%88%E7%B4%AF%E4%B9%98%E6%B3%95%EF%BC%89%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E4%B8%AD%2Ca1%3D1%2C%EF%BC%882n%2B1%EF%BC%89an%3D%EF%BC%882n-3%EF%BC%89an-1%EF%BC%88n%E2%89%A52%EF%BC%89%2C%E5%88%99%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
求数列的通项公式(累乘法)已知数列{an}中,a1=1,(2n+1)an=(2n-3)an-1(n≥2),则数列{an}的通项公式
求数列的通项公式(累乘法)
已知数列{an}中,a1=1,(2n+1)an=(2n-3)an-1(n≥2),则数列{an}的通项公式
求数列的通项公式(累乘法)已知数列{an}中,a1=1,(2n+1)an=(2n-3)an-1(n≥2),则数列{an}的通项公式
q=an/an-1=(2n-3)/(2n+1)
an/a1=a2/a1*a3/a2*a4/a3*……*an-1/an-2*an/an-1=1/5*3/7*5/9*……*(2n-5)/(2n-1)*(2n-3)/(2n+1)=3/{(2n-1)(2n+1)}
an=3/{(2n-1)(2n+1)}
代入检验n=1时也成立
所以an=3/{(2n-1)(2n+1)}(n≥1)
(2n+1)an=(2n-3)a(n-1)
∴an=(2n-3)/(2n+1)a(n-1)
=(2n-3)/(2n+1)×(2n-5)/(2n-1)a(n-2)
....
=[(2n-3)(2n-5)(2n-7)...1/(2n+1)(2n-1)(2n-3)(2n-5)....5]×a1
=(3×1)/(2...
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(2n+1)an=(2n-3)a(n-1)
∴an=(2n-3)/(2n+1)a(n-1)
=(2n-3)/(2n+1)×(2n-5)/(2n-1)a(n-2)
....
=[(2n-3)(2n-5)(2n-7)...1/(2n+1)(2n-1)(2n-3)(2n-5)....5]×a1
=(3×1)/(2n+1)(2n-1)
=3/(4n²-1)
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