数列{an}满足a1=1 an+1=2n+1an/an+2n
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数列{an}满足a1=1 an+1=2n+1an/an+2n
(1)a(n+1)/2^(n+1) =an/(an+2^n)
2^(n+1)/a(n+1)=(an+2^n)/an =1+2^n/an
2^(n+1)/a(n+1)-2^n/an=1
所以{2^n/an}是以公差d=1的等差数列
(2)2^n/an=a1+(n-1)d =1+n-1=n
所以an=2^n/n
(3)bn=n(n+1) an =(n+1)2^n
Sn=b1+b2+...+bn=2*2+3*2^2+4*2^3+.+(n+1)2^n
2Sn=2b1+2b2+.2bn
=2*2^2+3*2^3+4*2^4+...+n*2^n+(n+1)2^(n+1)
2Sn-Sn=(2*2^2-3*2^2)+(3*2^3-4*2^3)+.+(n*2^n-(n+1)2^n) +(n+1)2^(n+1)-2*2
=-[2^2+2^3+.+2^n]+(n+1)2^(n+1)-4 (下面就不求了)
再问: ����棬�мӷ�
再答: =-[2+2^2+2^3+...+2^n]+(n+1)*2^(n+1)-2 =- 2(1-2^n)/(1-2) +(n+1)*2^(n+1)-2 =2-2^(n+1)+(n+1)2^(n+1)-2 =n*2^(n+1)
2^(n+1)/a(n+1)=(an+2^n)/an =1+2^n/an
2^(n+1)/a(n+1)-2^n/an=1
所以{2^n/an}是以公差d=1的等差数列
(2)2^n/an=a1+(n-1)d =1+n-1=n
所以an=2^n/n
(3)bn=n(n+1) an =(n+1)2^n
Sn=b1+b2+...+bn=2*2+3*2^2+4*2^3+.+(n+1)2^n
2Sn=2b1+2b2+.2bn
=2*2^2+3*2^3+4*2^4+...+n*2^n+(n+1)2^(n+1)
2Sn-Sn=(2*2^2-3*2^2)+(3*2^3-4*2^3)+.+(n*2^n-(n+1)2^n) +(n+1)2^(n+1)-2*2
=-[2^2+2^3+.+2^n]+(n+1)2^(n+1)-4 (下面就不求了)
再问: ����棬�мӷ�
再答: =-[2+2^2+2^3+...+2^n]+(n+1)*2^(n+1)-2 =- 2(1-2^n)/(1-2) +(n+1)*2^(n+1)-2 =2-2^(n+1)+(n+1)2^(n+1)-2 =n*2^(n+1)
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