Question

数列{an}中,a1=1,Sn+1=4an+2设bn=an+1-2an,求证{bn}是等比数列,并求{an}通项.

Answer 1

S(n+1)-Sn=4(an-a(n-1))

即a(n+1)=4(an-a(n-1)) b(n+1)=a(n+1)-2an=2(an-2a(n-1))=2bn

既然你已经作出第一问,我就直接跳过

S2=4a1+2 a1+a2=4a1+2 a2=3a1+2=5

b1=a2-2a1=5-2=3 bn=3*2^(n-1)

a(n+1)-2an=3*2^(n-1) 两边同时除以2^(n+1)

a(n+1)/2^(n+1)-an/2^n=3/4

所以{an/2^n}成等差数列,公差3/4

an/2^n=1/2+3/4(n-1)=3/4n-1/4 an=3*[2^(n-2)]*n-2^(n-2)