Question

设数列an的前n项和为Sn,b∈R,且满足ban-2^n=(b-1)Sn.

Answer 1

a1=2

(b-1)Sn=ban - 2^n

(b-1)Sn-1=b(an-1) - 2^(n-1)

两式作差

an=b(an-1) + 2^(n-1)

两边同时除以2^n

得an/2^n=(b/2)*(an-1)/(2^n-1)+1/2

记cn=an/2^n

cn-1/(2-b)=(b/2)*[(cn-1)-1/2-b]

=[(b/2)^n-1]*(1-1/2-b)

an=2^n*[1/2-b+(b/2)^n-1*(1-b)/(2-b)]