已知数列{an}的前项和为sn,且sn+1=4an+2(n∈N+),a1=1,. (1)设bn=an+1-2an,求b1并证明数列{bn}为等比数列; (2)设cn=an2n,求证{cn}是等差数列.

问题描述:

已知数列{an}的前项和为sn,且sn+1=4an+2(n∈N+),a1=1,.
(1)设bn=an+1-2an,求b1并证明数列{bn}为等比数列;
(2)设cn=

an
2n
,求证{cn}是等差数列.

(1)∵a1=1,s2=4a1+2,得a2=s2-a1=3a1+2=5,∴b1=5-2=3,由sn+1=4an+2,得sn+2=4an+1+2,两式相减得sn+2-sn+1=4(an+1-an),即an+2=4(an+1-an),亦即an+2-2an+1=2an+1-4an∵bn=an+1-2an,∴bn+1=2bn∴bn+1bn=2...