绍兴市2023学年高二第二学期高中期末调测信息技术试题及答案

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19解:(1)当a=2时,Sn=n2+n+3当n=1时,a1=S1=5;当n≥2时,an=S。-Sn-1=2n经检验,a1=5不符合上式,故数列{an}的通项公式为an=(4分2n,n≥22)当n=1时,a1=S1=3+a;当n≥2时因为数列{an}是等差数列,所以3+a=2,解得a=-1.因为所以b2(n+1)2(n+1)n[(n+1)2+n+(n+1)(n+2)11nn+2故T,=b1+b2+…+bn53-2所以TS1=(n+1)(n+2)(一n+1n+23(n+1)(n+2)2(n+2)-(n+1)3(n+1)(n+2)2(2n+3)令TnSn+1=3(an+1).整理得3n2-7n-6=0,解得n=3或n=-、2(舍去)故存在n=3满足题意(12分)

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