同除以2^(n+1)
a(n+1)/2^(n+1)=an/2^n+(3/2)^(n+1)
则
an/2^n-a(n-1)/2^(n-1)=(3/2)^(n-1)
……
a2/2^2-a1/2^1=(3/2)^(1-1)
相加
an/2^n-a1/2=1+(3/2)+……+(3/2)^(n-1)=1*[1-(3/2)^n]/(1-3/2)=-2+2*(3/2)^n
所以an=2^n*[a1/2-2+2*(3/2)^n]
同除以2^(n+1)
a(n+1)/2^(n+1)=an/2^n+(3/2)^(n+1)
则
an/2^n-a(n-1)/2^(n-1)=(3/2)^(n-1)
……
a2/2^2-a1/2^1=(3/2)^(1-1)
相加
an/2^n-a1/2=1+(3/2)+……+(3/2)^(n-1)=1*[1-(3/2)^n]/(1-3/2)=-2+2*(3/2)^n
所以an=2^n*[a1/2-2+2*(3/2)^n]