设数列an的前n项和为Sn,a1=1,Sn+1=4an+2(n∈N+)
问题描述：
设数列an的前n项和为Sn,a1=1,Sn+1=4an+2(n∈N+)
1、若bn=an+1-2an,求bn
2、若cn=1/an+1-2an,求cn的前6项和T6
3、若dn=an/2^n,证明dn是等差数列
答
证明：(1)由于S(n+1)=4an+2则有:Sn=4a(n-1)+2两式相减,得:S(n+1)-Sn=4（an-a(n-1)）a(n+1)-2an=2[an-2a(n-1)]由于bn=A(n+1)-2an则有:bn=2b(n-1)则:bn/b(n-1)=2则:数列{bn}是公比为2的等比数列.(2)由于:bn=b1*2^(n-1)...