a(n+1)=1/2an+(1/2)^n
同除(1/2)^(n+1)得
a(n+1)*2^(n+1)=an*2^n+2
{an*2^n}是首项为2公差为2的等差数列
即an*2^n=2n得an=n/2^(n-1)
故bn=2^(n-1)*an=n 是等差数列
a(n+1)=1/2an+(1/2)^n
同除(1/2)^(n+1)得
a(n+1)*2^(n+1)=an*2^n+2
{an*2^n}是首项为2公差为2的等差数列
即an*2^n=2n得an=n/2^(n-1)
故bn=2^(n-1)*an=n 是等差数列