摘要：15．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)求{an}的通项公式, (2)设数列{bn}满足an(2bn-1)＝1.并记Tn为{bn}的前n项和.求证:3Tn+1>log2(an+3).n∈N*. (1)解:由a1＝S1＝(a1+1)(a1+2).解得a1＝1或a1＝2.由已知a1＝S1>1.因此a1＝2. 又由an+1＝Sn+1-Sn＝(an+1+1)(an+1+2)-(an+1)(an+2). 得(an+1+an)(an+1-an-3)＝0. 即an+1-an-3＝0或an+1＝-an.因an>0.故an+1＝-an不成立.舍去． 因此an+1-an＝3.从而{an}是公差为3.首项为2的等差数列.故{an}的通项为an＝3n-1. (2)证法一:由an(2bn-1)＝1可解得 bn＝log2＝log2, 从而Tn＝b1+b2+-+bn ＝log2. 因此3Tn+1-log2(an+3) ＝log2. 令f(n)＝3·. 则＝·3 ＝. 因(3n+3)3-(3n+5)(3n+2)2＝9n+7>0.故f(n+1)>f(n)． 特别地f(n)≥f(1)＝>1.从而3Tn+1-log2(an+3)＝log2f(n)>0.即3Tn+1>log2(an+3)． 证法二:同证法一求得bn及Tn. 由二项式定理知.当c>0时.不等式(1+c)3>1+3c成立． 由此不等式有 3Tn+1＝log22(1+)3(1+)3-(1+)3 >log22 ＝log22···-·＝log2(3n+2)＝log2(an+3)． 证法三:同证法一求得bn及Tn. 令An＝··-·.Bn＝··-·. Cn＝··-·. 因>>.因此A>AnBnCn＝. 从而3Tn+1＝log223＝log22A>log22AnBnCn＝log2(3n+2)＝log2(an+3)． 证法四:同证法一求得bn及Tn. 下面用数学归纳法证明:3Tn+1>log2(an+3)． 当n＝1时.3T1+1＝log2.log2(a1+3)＝log25. 因此3T1+1>log2(a1+3).结论成立． 假设结论当n＝k时成立.即3Tk+1>log2(ak+3). 则当n＝k+1时. 3Tk+1+1-log2(ak+1+3) ＝3Tk+1+3bk+1-log2(ak+1+3) >log2(ak+3)-log2(ak+1+3)+3bk+1 ＝log2. 因(3k+3)3-(3k+5)(3k+2)2＝9k+7>0.故 log2>0. 从而3Tk+1+1>log2(ak+1+3)．这就是说.当n＝k+1时结论也成立． 综上3Tn+1>log2(an+3)对任何n∈N*成立．

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