设数列{bn}的前n项和为Sn,且bn+Sn=n+2(n∈N*),数列{an}满足:a(1)=2/3,a(n+1)=bn
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设数列{bn}的前n项和为Sn,且bn+Sn=n+2(n∈N*),数列{an}满足:a(1)=2/3,a(n+1)=bn+1/n^2
(1)求数列bn的通项公式
(2)求证:2≤an<2/3e^2
(1)求数列bn的通项公式
(2)求证:2≤an<2/3e^2
(1)数列{bn}的前n项和为Sn,且bn+Sn=n+2(n∈N*),①
∴2b1=3,b1=3/2.
n>1时b+S=n+1,②
①-②,2bn-b=1,
变形得bn-1=(1/2)(b-1),
∴数列{bn-1}是等比数列,b1-1=1/2,公比=1/2,
bn-1=(1/2)^n,bn=1+1/2^n.
(2)a=bn+1/n^2 =1+1/2^n+1/n^2,
∴an=1+1/2^(n-1)+1/(n-1)^2,(n>1).
∴n→∞时,an→1,命题不成立.
∴2b1=3,b1=3/2.
n>1时b+S=n+1,②
①-②,2bn-b=1,
变形得bn-1=(1/2)(b-1),
∴数列{bn-1}是等比数列,b1-1=1/2,公比=1/2,
bn-1=(1/2)^n,bn=1+1/2^n.
(2)a=bn+1/n^2 =1+1/2^n+1/n^2,
∴an=1+1/2^(n-1)+1/(n-1)^2,(n>1).
∴n→∞时,an→1,命题不成立.
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