数列a1=1/2,a(n-1)+1=2an(n≥2)求数列An的通项公式若数列Bn满足：2b1+2^2b2+~+2^nbn=n2^n,求数列bn的通项公式令Cn=2An×Bn,求数列cn的前n项和Tn

Question

数列a1=1/2,a(n-1)+1=2an(n≥2)
求数列An的通项公式
若数列Bn满足：2b1+2^2b2+~+2^nbn=n2^n,求数列bn的通项公式
令Cn=2An×Bn,求数列cn的前n项和Tn

Answer 1

由a(n-1)+1=2an变形得
a(n-1)-1=2(an-1)
a(n-1)+1=2an
即an-1=(1/2)[a(n-1)-1]
所以数列{an-1}是以a1-1=1/2-1=-1/2为首项,1/2为公比的等比数列
于是an-1=(-1/2)(1/2)^(n-1)+1
即an=-(1/2)^n+1
2、
由2b1+2^2b2+~+2^nbn=n2^n得
2b1+2^2b2+~+2^(n-1)b(n-1)=(n-1)2^(n-1)
两式相减得2^nbn=n2^n-(n-1)2^(n-1)
2^nbn=(2n+1)2^(n-1)
即bn=(2n+1)/2
3、
Cn=2anbn=-（2n+1)(1/2)^n+(2n+1)=-(2n+1)/2^n+(2n+1)
设数列{(2n+1)/2^n}的前n项和为Sn,数列{2n+1}前n项和为Pn,则Tn=-Sn+Pn
Sn=3/2+5/2²+7/2³+.+(2n+1)/2^n
(1/2)Sn=3/2²+5/2³+.+(2n-1)/2^n+(2n+1)/2^(n+1)
上两式错项相减得（1/2)Sn=3/2+2/2²+2/2³+2/2^n-(2n+1)/2^(n+1)
(1/2)Sn=3/2+1/2+1/2²+1/2³+.+1/2^(n-1)-(2n+1)/2^(n+1)
(1/2)Sn=1/2+1+1/2+1/2²+1/2³+.+1/2^(n-1)-(2n+1)/2^(n+1)
(1/2)Sn=1/2+2[1-(1/2)^n]-(2n+1)/2^(n+1)
Sn=5-(2n+5)/2^n
Pn=3+5+7.+(2n+1)=n(n+1)
所以Tn=-Sn+pn=-5+(2n+5)/2^n+n(n+1)
方法是对的,看我算没算错,你自己检查一下.