Question

设数列{an}的前n项和Sn=2(an)-1,数列{bn}满足b1=3,bk+1=ak+bk

Answer 1

1.S(n-1)=2a(n-1)-1 S1=a1=2(a1)-1 a1=1

an=Sn-S(n-1)=2an-2a(n-1)

an=2a(n-1)

an=a1*2^(n-1)=2^(n-1)

2.b(k+1)=2^(k-1)+bk

bk=2^(k-2)+b(k-1)

bk/(2^k)=(1/2)*(b(k-1)/(2^(k-1)))+1/4