已知各项均为正数的数列{An}的前n项和Sn满足S1>1,且6Sn=(1+An)(2+An),n∈N*1) 求{An}的通项公式2) 设数列{Bn}满足An(-1+2^(Bn))=1,并记Tn为{bn}的前n项和,求证：3Tn+1>log(3+An)/log2注：第2问的‘3Tn+1>log(3+An)/log2’改为‘1+3Tn>log(3+An)/log2’防止误解~

1)
6Sn=An^2+3An+2
因为S1=A1
所以6A1=A1^2+3A1+2
A1^2-3A1+2=0
(A1-1)(A1-2)=0
因为A1=S1>1
所以A1=2
因为An=Sn-S(n-1) 注S(n-1)指前n-1项的和
An=(An^2+3An+2)/6-(A(n-1)^2+3A(n-1)+2)/6
所以6An=(An^2-A(n-1)^2+3(An-A(n-1))
An^2-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
因为{An}各项均为正数,所以An+A(n-1)>0
所以An-A(n-1)-3=0
即 An-A(n-1)=3
所以{An}是公差为3的等差数列
所以An=A1+(n-1)*3=2+3n-3
即 An=3n-1
2)
An(-1+2^(Bn))=1
即 2^(Bn)=1/An+1
Bn=log(1/An+1)/log2
因为An=3n-1
所以Bn=log(3n/(3n-1))/log2
Tn=B1+B2+……+Bn
=log(3/2)/log2+log(6/5)/log2+……+log(3n/(3n-1))/log2
=log[(3/2)*(6/5)*(9/8)*……*(3n/(3n-1)]/log2
因为算术平均值>=几何平均值>=调和平均值
所以(3/2)*(6/5)*(9/8)*……*(3n/(3n-1)
>={n/[(2/3)+(5/6)+……+(3n-1)/3n]}^n
={n/[n-(1/3+1/6+1/9+……+1/3n]}^n
>={n/[n-n*开n次根号下[(1/3)*(1/6)*(1/9)*……*(1/3n)]]}^n
={n/[n-n*开n次根号下[((1/3)^n)*(1/n!)]]}^n
={n/[n-(n/3)*开n次根号下(1/n!)]}^n
={1/[(2/3)*开n次根号下(1/n!)]}^n
={(3/2)*开n次根号下(n!)]}^n
=((3/2)^n)*(n!)
所以Tn>=log[((3/2)^n)*(n!)]/log2
所以1+3Tn>=log2/log2+3log[((3/2)^n)*(n!)]
=log2*((3/2)^3n)*[(n!)^3]/log2
而log(3+An)/log2=log(3n+2)/log2
而2*((3/2)^3n)*[(n!)^3]>(3n+2)
得证