Question

数列{an}满足a1=a,An+1=cAn＋1－cn∈N*，

Answer 1

1)a(n+1)-1=c(an-1)
所以{an}是以首项为a,公比为c的等比数列
an=a1*q^(n-1)=a*c^(n-1)
2)当a=1/2，c＝1/2
时，an=(1/2)^n
bn=n[1-(1/2)^n]
Sn=b1+b2+……+bn=(1+2+……+n)-[1*(1/2)+2*(1/2)^2+……+n*(1/2)^n]
令T=1*(1/2)+2*(1/2)^2+………………………………+n*(1/2)^n①
T/2=
1*(1/2)^2+2*(1/2)^3+……+（n-1)*(1/2)^n+n*(1/2)^(n+1)②
①-②得T/2=1/2+(1/2)^2+……+(1/2)^n-n*(1/2)^(n+1)=1-(1/2）^n-n*(1/2)^(n+1)
T=2-(n+2)*(1/2)^n
Sn=b1+b2+……+bn=(1+2+……+n)-[1*(1/2)+2*(1/2)^2+……+n*(1/2)^n]
=[n*(n+1)]/2+[(n+2)*(1/2)^n]-2

Answer 2

1)a(n+1)-1=c(an-1)
所以{an}是以首项为a,公比为c的等比数列
an=a1*q^(n-1)=a*c^(n-1)
2)当a=1/2，c＝1/2
时，an=(1/2)^n
bn=n[1-(1/2)^n]
Sn=b1+b2+……+bn=(1+2+……+n)-[1*(1/2)+2*(1/2)^2+……+n*(1/2)^n]
令T=1*(1/2)+2*(1/2)^2+………………………………+n*(1/2)^n①
T/2=
1*(1/2)^2+2*(1/2)^3+……+（n-1)*(1/2)^n+n*(1/2)^(n+1)②
①-②得T/2=1/2+(1/2)^2+……+(1/2)^n-n*(1/2)^(n+1)=1-(1/2）^n-n*(1/2)^(n+1)
T=2-(n+2)*(1/2)^n
Sn=b1+b2+……+bn=(1+2+……+n)-[1*(1/2)+2*(1/2)^2+……+n*(1/2)^n]
=[n*(n+1)]/2+[(n+2)*(1/2)^n]-2
恩
好了吧~~