已知數列{an}的前n項和為Sn,且S(n+1)=4an+2,a1=1,設Cn=an/2^n,求證數列{Cn}是等差數列

已知數列{an}的前n項和為Sn,且S(n+1)=4an+2,a1=1,設Cn=an/2^n,求證數列{Cn}是等差數列


S(n+1)=4an+2Sn=4a(n-1)+2a(n+1)=S(n+1)-Sn=4(an-a(n-1))a(n+1)-2an=2[an-2a(n-1)][a(n+1)-2an]/[an-2a(n-1)]=2{an-2a(n-1)}是公比為2的等差數列a1=1S2=4+2=6,a2=S2-a1=6-1=5an-2a(n-1)=[a2-2a1]*2^(n-2)=3* 2^(n-2)a…