设数列{an}前n项和为Sn,且(3-m)Sn+2man=m+3(n属于N*).其中m为实常数,m不等于-3且m不等于0.1,求证:{an}是等比数列.2,若数列{an}的公比满足q=f(m)且b1=a1,bn=3/2f(bn-1)(n属于N*,n大于等于2),求{bn}的通

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设数列{an}前n项和为Sn,且(3-m)Sn+2man=m+3(n属于N*).其中m为实常数,m不等于-3且m不等于0.1,求证:{an}是等比数列.2,若数列{an}的公比满足q=f(m)且b1=

设数列{an}前n项和为Sn,且(3-m)Sn+2man=m+3(n属于N*).其中m为实常数,m不等于-3且m不等于0.1,求证:{an}是等比数列.2,若数列{an}的公比满足q=f(m)且b1=a1,bn=3/2f(bn-1)(n属于N*,n大于等于2),求{bn}的通
设数列{an}前n项和为Sn,且(3-m)Sn+2man=m+3(n属于N*).其中m为实常数,m不等于-3且m不等于0.
1,求证:{an}是等比数列.
2,若数列{an}的公比满足q=f(m)且b1=a1,bn=3/2f(bn-1)(n属于N*,n大于等于2),求{bn}的通项公式.
3,若m=1时,设Tn=a1+2a2+3a3+...+nan(n属于N*),是否存在最大的正整数k,使得对任意n属于N*均有Tn大于k/8成立,若存在求出k的值,若不存在请说明理由.

设数列{an}前n项和为Sn,且(3-m)Sn+2man=m+3(n属于N*).其中m为实常数,m不等于-3且m不等于0.1,求证:{an}是等比数列.2,若数列{an}的公比满足q=f(m)且b1=a1,bn=3/2f(bn-1)(n属于N*,n大于等于2),求{bn}的通
1.
(3-m)Sn+2man=m+3 (1)
当n=1时,求得a1=1
当n=n-1时,
(3-m)S(n-1)+2ma(n-1)=m+3 (2)
第一式减去第二式得:
(3-m)an+2m(an-a(n-1))=0
即:an/a(n-1)=2m/(m+3)
所以{an}是等比数列,公比为2m/(m+3)
an=[2m/(m+3)]^(n-1)
2.
bn=3/2·2b(n-1)/[b(n-1)+3]=3b(n-1)/[b(n-1)+3]
所以:3/bn=1+3/b(n-1)
即:3/bn-3/b(n-1)=1
{3/bn}为等差数列,公差d=1,首项3/b1=3
3/bn=3+(n-1)=n+2
bn=3/(n+2)
3.
m=1,q=2m/(m+3)=0.5
an=(1/2)^(n-1)
Tn=a1+2a2+3a3+...+nan
0.5Tn=a2+2a3+3a4+……+(n-1)an+na(n+1)
0.5Tn=a1+a2+a3+……+an-na(n+1)
Tn=2[Sn-na(n+1)]=2Sn-2na(n+1)=2[2-(n+2)/2^n]
假设存在,Tn>k/8
k

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