已知数列an,a1=3,a(n+1)=2an -1 ,bn=2^n/an*a(n+1)(1)求证Sn
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已知数列an,a1=3,a(n+1)=2an -1 ,bn=2^n/an*a(n+1)(1)求证Sn<1/3,(2)比较an和n^2大小
![已知数列an,a1=3,a(n+1)=2an -1 ,bn=2^n/an*a(n+1)(1)求证Sn](/uploads/image/z/17731912-40-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%2Ca1%3D3%2Ca%EF%BC%88n%2B1%EF%BC%89%3D2an+-1+%2Cbn%3D2%5En%2Fan%2Aa%28n%2B1%29%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81Sn)
∵an+1=2an-1 ∴an+1-1=2(an-1) ∴an-1=(a1-1) ×2^(n-1)=2^n ∴an=2^n+1
∵bn=2^n/an×a(n+1)=2^n/﹛(2^n+1)[2^(n+1)+1]﹜
=1/(2^n+1)-1/[2^(n+1)+1]
(1)Sn=(1/3-1/5)+(1/5-1/9)+···+﹛1/(2^n+1)-1/[2^(n+1)+1]﹜
=1/3-1/[2^(n+1)+1]
∵1/[2^(n+1)+1]>0 ∴Sn<1/3
(2)n=1,a1=3>1²
n=2,a2=5>2²
n=3,a3=9=3²
n=4,a4=17>4²
n=5,a5=33>5²
∴an≥n²(当n=3时,取=)
∵bn=2^n/an×a(n+1)=2^n/﹛(2^n+1)[2^(n+1)+1]﹜
=1/(2^n+1)-1/[2^(n+1)+1]
(1)Sn=(1/3-1/5)+(1/5-1/9)+···+﹛1/(2^n+1)-1/[2^(n+1)+1]﹜
=1/3-1/[2^(n+1)+1]
∵1/[2^(n+1)+1]>0 ∴Sn<1/3
(2)n=1,a1=3>1²
n=2,a2=5>2²
n=3,a3=9=3²
n=4,a4=17>4²
n=5,a5=33>5²
∴an≥n²(当n=3时,取=)
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