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已知数列{an}的前n项和为Sn,a1=1,且满足anan+1=2Sn,数列{bn}满足b1=16,bn+1-bn=2n,则数列{bnan}中第项最小.
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已知数列{an}的前n项和为Sn,a1=1,且满足anan+1=2Sn,数列{bn}满足b1=16,bn+1-bn=2n,则数列{
}中第___项最小.
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答案和解析
当n=1时,2S1=a1a2,即2a1=a1a2,∴a2=2.
当n≥2时,2Sn=anan+1,2Sn-1=an-1an,两式相减得2an=an(an+1-an-1),
∵an≠0,∴an+1-an-1=2,
∴{a2k-1},{a2k}都是公差为2的等差数列,又a1=1,a2=2,
∴{an}是公差为1的等差数列,
∴an=1+(n-1)×1=n,
∵b1=16,bn+1-bn=2n,∴bn =( bn -bn-1)+( bn-1 -bn-2)+ ( bn-2 -bn-3)+…+( b2 -b1)+b1=n(n-1)+16
=n+
-1,利用基本不等式得n=4时n+
-1最小,∴数列{
}中第 4项最小.
当n≥2时,2Sn=anan+1,2Sn-1=an-1an,两式相减得2an=an(an+1-an-1),
∵an≠0,∴an+1-an-1=2,
∴{a2k-1},{a2k}都是公差为2的等差数列,又a1=1,a2=2,
∴{an}是公差为1的等差数列,
∴an=1+(n-1)×1=n,
∵b1=16,bn+1-bn=2n,∴bn =( bn -bn-1)+( bn-1 -bn-2)+ ( bn-2 -bn-3)+…+( b2 -b1)+b1=n(n-1)+16
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