数列前n项和为Sn,a1=1,S（n +1）=4(an)+2.设bn=an/(2的n次方),证:(bn)是等差数列.

S（n +1）=4(an)+2
Sn =4(an-1)+2
两式相减
(an+1)=4(an)-4(an-1)
(an+1)+4(an-1)=4(an)
(an+1)/2^(n+1)+4(an-1)/2(n+1)=4(an)/2^(n+1)
(bn+1)+(bn-1)=2(bn)
(bn)是首项为1/2的等差数列