已知数列{an}的前n项和Sn,且a1=a(a为非零常数),当n>=2时,an=2Sn^2/2
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已知数列{an}的前n项和Sn,且a1=a(a为非零常数),当n>=2时,an=2Sn^2/2
已知数列{an}的前n项和Sn,且a1=a(a为非零常数),当n>=2时,an=2Sn^2/2Sn-1
1)求证:数列{1/Sn}是等差数列
2)设bn=Sn/an,数列bn的前n项和为Tn.已知当且仅当n=6时,Tn有最大值,求实数a取值范围.
已知数列{an}的前n项和Sn,且a1=a(a为非零常数),当n>=2时,an=2Sn^2/2Sn-1
1)求证:数列{1/Sn}是等差数列
2)设bn=Sn/an,数列bn的前n项和为Tn.已知当且仅当n=6时,Tn有最大值,求实数a取值范围.
n>=2,
an=Sn-Sn-1=2Sn^2/2Sn-1
==>1/Sn-1/Sn-1=2
==>{1/Sn}是以2为公差的等差数列
Sn=a/[1+(2n-2)*a]
==>bn=Sn/an=-n+2-1/(2a)
==>Tn=(-1/2)*n^2+[(3a-1)/(2a)]*n
==>对称轴n=(3a-1)/(2a)在区间(5.5,6.5)
==>-1/8
an=Sn-Sn-1=2Sn^2/2Sn-1
==>1/Sn-1/Sn-1=2
==>{1/Sn}是以2为公差的等差数列
Sn=a/[1+(2n-2)*a]
==>bn=Sn/an=-n+2-1/(2a)
==>Tn=(-1/2)*n^2+[(3a-1)/(2a)]*n
==>对称轴n=(3a-1)/(2a)在区间(5.5,6.5)
==>-1/8
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