已知各项均为正数的数列{an}的前n项和Sn满足S1>1,且6Sn=(an+1)(an+2),n∈N+1.设数列{bn}满足an((2^bn)-1)=1,并记Tn为{bn}的前n项和,求证：3Tn+1>log2(an+3),n∈N+.请一定要详细作答an((2^bn)-1)=1为an乘(2的bn次方-1

已知各项均为正数的数列{an}的前n项和Sn满足S1>1,且6Sn=(an+1)(an+2),n∈N+1.设数列{bn}满足an((2^bn)-1)=1,并记Tn为{bn}的前n项和,求证：3Tn+1>log2(an+3),n∈N+.请一定要详细作答an((2^bn)-1)=1为an乘(2的bn次方-1
已知各项均为正数的数列{an}的前n项和Sn满足S1>1,且6Sn=(an+1)(an+2),n∈N+
1.设数列{bn}满足an((2^bn)-1)=1,并记Tn为{bn}的前n项和,求证：3Tn+1>log2(an+3),n∈N+.
请一定要详细作答
an((2^bn)-1)=1为an乘(2的bn次方-1的差)=1
log2(an+3)为以2为底,an+3为真数的对数.
555Alan555的看不懂.麻烦详解.

已知各项均为正数的数列{an}的前n项和Sn满足S1>1,且6Sn=(an+1)(an+2),n∈N+1.设数列{bn}满足an((2^bn)-1)=1,并记Tn为{bn}的前n项和,求证：3Tn+1>log2(an+3),n∈N+.请一定要详细作答an((2^bn)-1)=1为an乘(2的bn次方-1
6S（n+1）-6Sn=6a（n+1）=【a（n+1）+1】【a（n+1）+2】-（an+1）（an+2）.
化简得【a（n+1）+an】【a（n+1）-an-3】=0.
因各项均为正数,故a（n+1）-an-3=0,即a（n+1）-an=3.
6S1=（a1+1）（a1+2）=6a1.解得a1=1或a1=2
因S1=a1>1,故a1=2.
故an=3n-1.

6Sn=(an+1)(an+2)①
6S(n-1)=(a(n-1)+1)(a(n-1)+2)②
①-② 整理得 an-a(n-1)=3
6S1=an^2+3an+2 得a1=2 an=3n-1
an((2^bn)-1)=1 看不懂，麻烦做详细说明

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