数列(an)满足a1=1,且8a(n+1)an -16a(n+1) +2an+5=0 ,记bn=1/an-0.5(n大于
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数列(an)满足a1=1,且8a(n+1)an -16a(n+1) +2an+5=0 ,记bn=1/an-0.5(n大于等于1)求数列(anbn)的通项公式及前n项和Sn (注:an为a的第n项,a(n+1)为a的第n+1项)
令anbn=cn,则cn=2an/(2an-1)
下面构造2an-1,2a(n+1)-1:
c(n+1)-cn
=a(n+1)/[2a(n+1)-1]-an/(2an-1)
={(2an-1)-[2a(n+1)-1]}/{2[2a(n+1)-1](2an-1)}
=1/{2[2a(n+1)-1]}-1/[2(2an-1)]
cn=1/[2(2an-1)]
c1=0.5
8a(n+1)an -16a(n+1) +2an+5=0
2(2an-1)[2a(n+1)-1]+3(2an-1)-6[2a(n+1)-1]=0
2(2an-1)[2a(n+1)-1]=3{2[2a(n+1)-1]-(2an-1)}
1/(2an-1)-1/{2[2a(n+1)-1]}=3
2cn-c(n+1)=3
c(n+1)=2cn-3
cn=2c(n-1)-3
cn=2c(n-1)-3
=2[2c(n-2)-3]-3
=2^2c(n-2)-3-3*2
=2^2[2c(n-3)-3]-3-3*2
=2^3c(n-3)- 3(1+2+2^2)
=...
=2^(n-1)c1-3[1+2+2^2+...+2^(n-2)]
=2^(n-2)-3[2^(n-1)-1]
=3-2.5*2^(n-1)
Sn=3n-2.5*[1+2+...+2^(n-1)]
=3n-2.5*(2^n-1)
=3n-2.5*2^n+2.5
下面构造2an-1,2a(n+1)-1:
c(n+1)-cn
=a(n+1)/[2a(n+1)-1]-an/(2an-1)
={(2an-1)-[2a(n+1)-1]}/{2[2a(n+1)-1](2an-1)}
=1/{2[2a(n+1)-1]}-1/[2(2an-1)]
cn=1/[2(2an-1)]
c1=0.5
8a(n+1)an -16a(n+1) +2an+5=0
2(2an-1)[2a(n+1)-1]+3(2an-1)-6[2a(n+1)-1]=0
2(2an-1)[2a(n+1)-1]=3{2[2a(n+1)-1]-(2an-1)}
1/(2an-1)-1/{2[2a(n+1)-1]}=3
2cn-c(n+1)=3
c(n+1)=2cn-3
cn=2c(n-1)-3
cn=2c(n-1)-3
=2[2c(n-2)-3]-3
=2^2c(n-2)-3-3*2
=2^2[2c(n-3)-3]-3-3*2
=2^3c(n-3)- 3(1+2+2^2)
=...
=2^(n-1)c1-3[1+2+2^2+...+2^(n-2)]
=2^(n-2)-3[2^(n-1)-1]
=3-2.5*2^(n-1)
Sn=3n-2.5*[1+2+...+2^(n-1)]
=3n-2.5*(2^n-1)
=3n-2.5*2^n+2.5
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