数列 a(n+1)=2an+1/(2^n),a1=2,求an

a(n+1)=2a(n)+1/(2^n)
a(n+1)+(1/3)*1/2^n)=2[a(n)+(1/3)*1/2^(n-1)]
即
a(n)+(1/3)*1/2^(n-1)=2^(n-1)*[a(1)+1/3]=(7/3)*2^(n-1),(n≥2)
所以
a(n)=[7*2^(n-1)-1/2^(n-1)]/3,(n≥2)
a(1)=2满足上式，所以
a(n)=[7*2^(n-1)-1/2^(n-1)]/3

a(n+1)=1/2an+1/2^n
an=1/2a(n-1)+1/2^(n-1)
=1/2[1/2a(n-2)+1/2^(n-2)]+1/2^(n-1)
=1/4a(n-2)+2*1/2^(n-1)
.....................
=1/2^(n-1)*a1+(n-1)*1/2^(n-1)
=n/2^(n-1)