2a(n+2)=an+a(n+1)等式俩边同时减去2a(n+1)∴2[a(n+2)-a(n+1)]=an-a(n+1)=-[a(n+1)-an]可知a(n+1)-an是以a2-a1=1为首项,以-1/2为公比的等比数列∴a(n+1)-an=(-1/2)^(n-1)∴an-a(n-1)=(-1/2)^(n-2),a(n-1)-a(n-1)=(-1...证明an的没怎么懂、、、