数列{an}满足a1=2,an+1=2^(n+1)an/((n+1/2)an+2^n),(1)设bn=2^n/an,求bn(2)设Cn=1/n（n+1）an+1数列{an}满足a1=2,an+1=2^(n+1)an/((n+1/2)an+2^n),（1）设bn=2^n/an,求bn（2）cn=1/n（n+1）an,数列{cn}的n项和Sn,求出Sn,并由此证

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$数列{an}满足a1=2,an+1=2^(n+1)*an/((n+1/2)*an+2^n),(1)设bn=2^n/an,求bn(2)设Cn=1/n（n+1）an+1数列{an}满足a1=2,an+1=2^(n+1)*an/((n+1/2)*an+2^n),（1）设bn=2^n/an,求bn（2）cn=1/n（n+1）an,数列{cn}的n项和Sn,求出Sn,并由此证$

数列{an}满足a1=2,an+1=2^(n+1)*an/((n+1/2)*an+2^n),(1)设bn=2^n/an,求bn(2)设Cn=1/n（n+1）an+1数列{an}满足a1=2,an+1=2^(n+1)*an/((n+1/2)*an+2^n),（1）设bn=2^n/an,求bn（2）cn=1/n（n+1）an,数列{cn}的n项和Sn,求出Sn,并由此证
数列{an}满足a1=2,an+1=2^(n+1)*an/((n+1/2)*an+2^n),(1)设bn=2^n/an,求bn(2)设Cn=1/n（n+1）an+1
数列{an}满足a1=2,an+1=2^(n+1)*an/((n+1/2)*an+2^n),（1）设bn=2^n/an,求bn
（2）cn=1/n（n+1）an,数列{cn}的n项和Sn,求出Sn,并由此证明5/16≤Sn<1/2

数列{an}满足a1=2,an+1=2^(n+1)*an/((n+1/2)*an+2^n),(1)设bn=2^n/an,求bn(2)设Cn=1/n（n+1）an+1数列{an}满足a1=2,an+1=2^(n+1)*an/((n+1/2)*an+2^n),（1）设bn=2^n/an,求bn（2）cn=1/n（n+1）an,数列{cn}的n项和Sn,求出Sn,并由此证
(1)2^(n+1)/a(n+1)=[(n+1/2)an+2^n]/an
即b(n+1)=bn+n+1/2
bn=b(n-1)+(n-1)+1/2
...
b2=b1+1+1/2
所以bn=(bn-b(n-1))+...+(b2-b1)+b1
=((n-1)+1/2)+...+(1+1/2)+1
=(1+2+...+(n-1))+(1/2+1/2+...+1/2)+1
=n(n-1)/2+(n-1)/2+1
=(n^2+1)/2
(2)an=2^n/bn=2^(n+1)/(n^2+1)
cn=(n^2+2n+2)/[n(n+1)2^(n+2)]=1/2{(n+1)/(n*2^n)-(n+2)/[(n+1)2^(n+1)]}
所以Sn=1/2[2/(1*2^1)-3/(2*2^2)]+1/2[3/(2*2^2)-4/(3*2^3)]+...+1/2{(n+1)/(n*2^n)-(n+2)/[(n+1)2^(n+1)]}
=1/2{2/(1*2^1)-(n+2)/[(n+1)2^(n+1)]}（裂项相消法）
=1/2-(n+2)/[(n+1)2^(n+2)]
首先Sn=1/2-(n+2)/[(n+1)2^(n+2)]0,即Sn单增
所以Sn≥S1=5/16

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