第1个回答 2011-06-09
(2n+3)a(n+1)=(2n-1)a(n)
(2n+1)a(n)=(2n-3)a(n-1)
(2n-1)a(n-1)=(2n-5)a(n-2)
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(2*3+3)a(3+1)=(2*5-1)a(4)=(2*3-1)a(3)
(2*2+3)a(2+1)=(2*4-1)a(3)=(2*2-1)a(2)
(2*1+3)a(1+1)=(2*3-1)a(2)=(2*1-1)a(1)
[(2n+3)(2n+1)(2n-1)...(2*3+3)(2*2+3)(2*1+3)]a(n+1)a(n)a(n-1)...a(4)a(3)a(2)=[(2n-1)(2n-3)(2n-5)...(2*3-1)(2*2-1)(2*1-1)]a(n)a(n-1)a(n-2)...a(3)a(2)a(1)
(2n+3)(2n+1)a(n+1)=(2*2-1)(2*1-1)a(1)=3
a(n+1)=3/[(2n+3)(2n+1)]
a(n)=3/[(2n+1)(2n-1)]
第2个回答 2011-06-09
a(n+1)=(2n-1)a(n)
a2=a1/5=1/5
a3=3a2/7=(3/7)*(1/5)
a4=(5/9)*(3/7)*(1/5)
a5=(7/11)*(5/9)*(3/7)*(1/5)
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an=(1/(2n+3))*(1/(2n+1))*3=3/(4n^2+8n+3)本回答被提问者采纳
第3个回答 2011-06-09
(2n+1)an=(2n-3)an-1
a1=1
a2=1/5
a3=(1/5)(3/7)
a4=(1/5)(3/7)(5/9)
n>=2
an=(1*3)/[(2n+1)(2n-1) ]