设数列an中的前n项的和为Sn,并且a1=1,Sn+1=4an+2.设bn=A(n+1)-2an,求证bn是等比数列

问题描述:

设数列an中的前n项的和为Sn,并且a1=1,Sn+1=4an+2.设bn=A(n+1)-2an,求证bn是等比数列

S2=4a1+2
a2=5
S(n+1)=4an+2
S(n+2)=4a(n+1)+2
a(n+2)=4a(n+1)-4an
a(n+2)-2a(n+1)=2(a(n+1)-2an)
b(n+1)=2bn
b1=a2-2a1=3
得证