摘要：2．若数列{an}的通项公式为an＝(1+n(1))n.试证:(1)数列{an}为递增数列,(2)2≤an＜3.证明:(1)an＝(1+n(1))n＝1+Cn(1)n(1)+Cn(2)(n(1))2+…+Cn(n)(n(1))n.an+1＝(1+n+1(1))n+1＝1+Cn+1(1)n+1(1)+Cn+1(2)(n+1(1))2+…+Cn+1(n+1)(n+1(1))n+1.可观察Cn+1(k)(n+1(1))k与Cn(k)(n(1))k.当k＝0,1时.Cn+1(k)(n+1(1))k＝Cn(k)(n(1))k,当k＝2,3,4.….n时.Cn+1(k)(n+1(1))k＞Cn(k)(n(1))k.∴an＜an+1.即{an}为递增数列．(2)∵an＝(1+n(1))n＝1+Cn(1)n(1)+Cn(2)(n(1))2+…+Cn(n)(n(1))n≥1+Cn(1)n(1)＝2.又an＝(1+n(1))n＝1+Cn(1)n(1)+Cn(2)(n(1))2+…+Cn(n)(n(1))n≤2+1×2(1)+2×3(1)+…+n(1)＝3-n(1)＜3.

网址：http://m.1010jiajiao.com/timu_id_3826945[举报]