数列a1+a2+...+an-1=(an-1)*2+(n-1)/2求(1)a1、a2、a3(2)an的通项公式数列a1+a2+...+a(n-1)=(an-1)*2/2+(n-1)/2求(1)a1、a2、a3(2)an的通项公式(3)bn=a(n+2)/[an*a(n+1)*2^(n-1)]求bn的前n项和tn
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![数列a1+a2+...+an-1=(an-1)*2+(n-1)/2求(1)a1、a2、a3(2)an的通项公式数列a1+a2+...+a(n-1)=(an-1)*2/2+(n-1)/2求(1)a1、a2、a3(2)an的通项公式(3)bn=a(n+2)/[an*a(n+1)*2^(n-1)]求bn的前n项和tn](/uploads/image/z/5367486-30-6.jpg?t=%E6%95%B0%E5%88%97a1%2Ba2%2B...%2Ban-1%3D%28an-1%29%2A2%2B%28n-1%29%2F2%E6%B1%82%EF%BC%881%EF%BC%89a1%E3%80%81a2%E3%80%81a3%EF%BC%882%EF%BC%89an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E6%95%B0%E5%88%97a1%2Ba2%2B...%2Ba%28n-1%29%3D%28an-1%29%2A2%2F2%2B%28n-1%29%2F2%E6%B1%82%EF%BC%881%EF%BC%89a1%E3%80%81a2%E3%80%81a3%EF%BC%882%EF%BC%89an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%283%29bn%3Da%28n%2B2%29%2F%5Ban%2Aa%28n%2B1%29%2A2%5E%28n-1%29%5D%E6%B1%82bn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8Ctn)
数列a1+a2+...+an-1=(an-1)*2+(n-1)/2求(1)a1、a2、a3(2)an的通项公式数列a1+a2+...+a(n-1)=(an-1)*2/2+(n-1)/2求(1)a1、a2、a3(2)an的通项公式(3)bn=a(n+2)/[an*a(n+1)*2^(n-1)]求bn的前n项和tn
数列a1+a2+...+an-1=(an-1)*2+(n-1)/2求(1)a1、a2、a3(2)an的通项公式
数列a1+a2+...+a(n-1)=(an-1)*2/2+(n-1)/2求(1)a1、a2、a3(2)an的通项公式(3)bn=a(n+2)/[an*a(n+1)*2^(n-1)]求bn的前n项和tn
数列a1+a2+...+an-1=(an-1)*2+(n-1)/2求(1)a1、a2、a3(2)an的通项公式数列a1+a2+...+a(n-1)=(an-1)*2/2+(n-1)/2求(1)a1、a2、a3(2)an的通项公式(3)bn=a(n+2)/[an*a(n+1)*2^(n-1)]求bn的前n项和tn
a1=(a1-1)*2+0
a1=2
a1+a2=2(a2-1)+(2-1)/2
2+a2=2a1-2+1/2
a2=7/2
a1+a2+a3=2(a3-1)+(3-1)/2
2+7/2+a3=2a3-2+1
a3=13/2
a1+a2+...+a(n-1)=S(n-1)=2(an-1)+(n-1)/2
Sn=2(a(n+1)-1)+n/2
下式-上式得到:an=2(a(n+1)-an)+1/2
2a(n+1)=3an-1/2
2(a(n+1)-1/2)=3(an-1/2)
故数列an-1/2是以首项是a1-1/2=3/2公比是3/2的等比数列,即有:an-1/2=3/2*(3/2)^(n-1)=(3/2)^n
即有an=(3/2)^2+1/2