Question

已知数列An满足a1=a(a为常数,a∈R）,a(n+1)=2^n-3an(n∈N*),设bn=an/2^n(n∈N*)

Answer 1

a1=a,a(n+1)=2^n-3an,两端除以2^（n+1）,

得a(n+1)/[2^(n+1)]=1/2-3*an/[2^(n+1)].

代入bn=an/2^n,得b(n+1)=1/2-3bn/2,

变形得(b(n+1)-0.5)=(-3/2)(bn-0.5);

所以数列cn=bn-0.5为等比数列,c1=b1-0.5=a/2-0.5;

所以cn=(-3/2)^n*(a/2-0.5);

得bn=(-3/2)^n*(a/2-0.5)-0.5;

an=bn/2^n＝(-3)^n*(a/2-0.5)/[2^(2n)]-2^[-(n+1)];