典型例题分析1: 已知数列{an}满足a1=5,a2=13,an+2=5an+1﹣6an,则使该数列的n项和Sn不小于2016的最小自然数n等于. 解:∵an+2=5an+1﹣6an, ∴an+2﹣2an+1=3(an+1﹣2an), an+2﹣3an+1=2(an+1﹣3an), 又∵a2﹣2a1=13﹣10=3,a2﹣3a1=13﹣15=﹣2, ∴数列{an+1﹣2an}是以3为首项,3为公比的等比数列, 数列{an+1﹣3an}是以﹣2为首项,2为公比的等比数列, ∴an+1﹣2an=3n,an+1﹣3an=﹣2n, ∴an=3n+2n,a1=5也成立; 故Sn=(3+2)+(4+9)+…+(3n+2n) =3(1-3n)/(1-3)+2(1-2n)/(1-2) =3(3n﹣1)/2+2(2n﹣1)≥2016, 故n≥7, 故答案为:7. 考点分析: 数列递推式. 题干分析: 化简an+2=5an+1﹣6an可得an+2﹣2an+1=3(an+1﹣2an),an+2﹣3an+1=2(an+1﹣3an),从而可知数列{an+1﹣2an},{an+1﹣3an}成等比数列,从而求得. |
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