设{an}是首项为1的正项数列,且(n+1)a(n+1)^2-nan^2+ana(n+1)=0,(n∈N*),求它的通项
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设{an}是首项为1的正项数列,且(n+1)a(n+1)^2-nan^2+ana(n+1)=0,(n∈N*),求它的通项公式(要迭代法)
(n+1)a(n+1)^2-nan^2+ana(n+1)=0
[(n+1)a(n+1)-nan] [ a(n+1) + an] =0
(n+1)a(n+1)-nan=0 or a(n+1)+an=0 (rejected)
a(n+1)/an = n/(n+1)
an/a(n-1) = (n-1)/n
an/a1 = 1/n
an= 1/n
再问: 谢谢啊,答案对了,可是我不明白an/a(n-1) = (n-1)/n怎么变成an/a1 = 1/n的
再答: an/a(n-1) = (n-1)/n [an/a(n-1)]. [a(n-1)/a(n-2)].....[a2/a1] =[(n-1)/n].[(n-2)/(n-1)]....[1/2] an/a1= 1/n an = a1(1/n) =1/n
[(n+1)a(n+1)-nan] [ a(n+1) + an] =0
(n+1)a(n+1)-nan=0 or a(n+1)+an=0 (rejected)
a(n+1)/an = n/(n+1)
an/a(n-1) = (n-1)/n
an/a1 = 1/n
an= 1/n
再问: 谢谢啊,答案对了,可是我不明白an/a(n-1) = (n-1)/n怎么变成an/a1 = 1/n的
再答: an/a(n-1) = (n-1)/n [an/a(n-1)]. [a(n-1)/a(n-2)].....[a2/a1] =[(n-1)/n].[(n-2)/(n-1)]....[1/2] an/a1= 1/n an = a1(1/n) =1/n
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