Question

等比数列{an}的前n项和sn=a*2^n+a-2,则an=?

Answer 1

Sn=a*2^n+a-2,

Sn-1=a*2^n-1+a-2,

Sn-Sn-1=an=a*2^n-a*2^n-1=a*2^(n-1)

S1=a1=3a-2

S2=a1+a2=5a-2 a2=2a

S3=a1+a2+a3=9a-2 a3=4a

(2a)^2=4a*(3a-2) 解得 a=1

an=2^(n-1)