已知各项均大于1的等差数列{an}前n项和为Sn且满足6Sn=an2+3an+2(n∈N+),数列{bn}满足bn=1/

1.
n=1时,6S1=6a1=a1²+3an+2
a1²-3a1+2=0
(a1-1)(a1-2)=0
a1=1(数列各项均>1,舍去)或a1=2
n≥2时,
6Sn=an²+3an+2
6S(n-1)=a(n-1)²+3a(n-1)+2
6Sn-6S(n-1)=6an=an²+3an+2-a(n-1)²-3a(n-1)-2
an²-a(n-1)²-3an-3a(n-1)=0
[an+a(n-1)][an-a(n-1)]-3[an+a(n-1)]=0
[an+a(n-1)][an-a(n-1)-3]=0
数列各项均>1 an+a(n-1)>2>0,要等式成立,只有an-a(n-1)=3,为定值.
又a1=2,数列{an}是以2为首项,3为公差的等差数列.
2.
an=2+3(n-1)=3n-1
bn=1/[ana(n+1)]=1/[(3n-1)(3n+2)]=(1/3)[1/(3n-1)-1/[3(n+1)-1]]
Tn=b1+b2+...+bn
=(1/3)[1/(3×1-1)-1/(3×2-1)+1/(3×2-1)-1/(3×3-1)+...+1/(3n-1)-1/(3(n+1)-1)]
=(1/3)[1/2 -1/(3n+2)]
=n/[2(3n+2)]
3.
假设存在满足题意的m,n,则有
2Tm=T2+Tn
2m/[2(3m+2)]=1/8 +n/[2(3n+2)]
整理,得
(3n+10)m=14n+4
m=(14n+4)/(3n+10)
要m为正整数,(14n+4)/(3n+10)为正整数.
令(14n+4)/(3n+10)=k (k为正整数)
(14-3k)n=10k-4
n=(10k-4)/(14-3k)=2(5k-2)/(14-3k),分子是偶数,要n为整数,分母同样为偶数,k为正偶数.
k≥1,5k-2≥3>0,要n为正数,14-3k≥0 k≤14/3,又k为正偶数,k只能是2,4
k=2时,n=(20-4)/8=2 与已知n>2矛盾,舍去
k=4时,n=(40-4)/(14-12)=18,此时m=(14n+4)/(3n+10)=(14×18+4)/(3×18+10)=4
m=n=4,又已知m