设数列an是首项为1的正项数列,且(n+1)a²n+1-na²n+an+1an=0(n=1,2,3.)求次数列的通项公式说的是先化简an与an+1的关系,为什么整理出来是[(n+1)an+1-nan](an+1+an)=0就是问
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![设数列an是首项为1的正项数列,且(n+1)a²n+1-na²n+an+1an=0(n=1,2,3.)求次数列的通项公式说的是先化简an与an+1的关系,为什么整理出来是[(n+1)an+1-nan](an+1+an)=0就是问](/uploads/image/z/4494699-27-9.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97an%E6%98%AF%E9%A6%96%E9%A1%B9%E4%B8%BA1%E7%9A%84%E6%AD%A3%E9%A1%B9%E6%95%B0%E5%88%97%2C%E4%B8%94%EF%BC%88n%EF%BC%8B1%EF%BC%89a%26%23178%3Bn%EF%BC%8B1%EF%BC%8Dna%26%23178%3Bn%EF%BC%8Ban%EF%BC%8B1an%3D0%EF%BC%88n%3D1%2C2%2C3.%EF%BC%89%E6%B1%82%E6%AC%A1%E6%95%B0%E5%88%97%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E8%AF%B4%E7%9A%84%E6%98%AF%E5%85%88%E5%8C%96%E7%AE%80an%E4%B8%8Ean%EF%BC%8B1%E7%9A%84%E5%85%B3%E7%B3%BB%2C%E4%B8%BA%E4%BB%80%E4%B9%88%E6%95%B4%E7%90%86%E5%87%BA%E6%9D%A5%E6%98%AF%5B%EF%BC%88n%EF%BC%8B1%EF%BC%89an%EF%BC%8B1%EF%BC%8Dnan%5D%EF%BC%88an%EF%BC%8B1%EF%BC%8Ban%EF%BC%89%3D0%E5%B0%B1%E6%98%AF%E9%97%AE)
设数列an是首项为1的正项数列,且(n+1)a²n+1-na²n+an+1an=0(n=1,2,3.)求次数列的通项公式说的是先化简an与an+1的关系,为什么整理出来是[(n+1)an+1-nan](an+1+an)=0就是问
设数列an是首项为1的正项数列,且(n+1)a²n+1-na²n+an+1an=0(n=1,2,3.)
求次数列的通项公式
说的是先化简an与an+1的关系,为什么整理出来是[(n+1)an+1-nan](an+1+an)=0
就是问上面一部是怎么得出来的
设数列an是首项为1的正项数列,且(n+1)a²n+1-na²n+an+1an=0(n=1,2,3.)求次数列的通项公式说的是先化简an与an+1的关系,为什么整理出来是[(n+1)an+1-nan](an+1+an)=0就是问
n[a(n+1)]²+[a(n+1)]²-n(an)²+a(n+1)an=0
n{[a(n+1)]²-(an)²}+[a(n+1)]²+a(n+1)an=0
n[a(n+1)+an][a(n+1)-an]+a(n+1)[a(n+1)+an]=0
[a(n+1)+an]{n[a(n+1)-an]+a(n+1)}=0
[a(n+1)+an][na(n+1)-nan+a(n+1)]=0
(n+1)a²n+1-na²n+an+1an=0
(n+1)*a(n+1)^2-n*an^2+an*a(n+1)=0
n*(a(n+1)^2-an^2)+a(n+1)^2+an*a(n+1)=0
(a(n+1)+an)((n+1)*a(n+1)-n*an)=0
又{an}为正项数列,(n+1)*a(n+1)-n*an=0
(n+1)*a(n+1)=n*an
1*a1=1
an=1/n
n[a(n+1)]²+[a(n+1)]²-n(an)²+a(n+1)an=0n{[a(n+1)]²-(an)²}+[a(n+1)]²+a(n+1)an=0n[a(n+1)+an][a(n+1)-an]+a(n+1)[a(n+1)+an]=0[a(n+1)+an]{n[a(n+1)-an]+a(n+1)}=0[a(n+1)+an][na(n...
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n[a(n+1)]²+[a(n+1)]²-n(an)²+a(n+1)an=0n{[a(n+1)]²-(an)²}+[a(n+1)]²+a(n+1)an=0n[a(n+1)+an][a(n+1)-an]+a(n+1)[a(n+1)+an]=0[a(n+1)+an]{n[a(n+1)-an]+a(n+1)}=0[a(n+1)+an][na(n+1)-nan+a(n+1)]=0
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