解；当n>=2时a(n+1)=snan=s(n-1)a(n+1)-an=an∴2an=a(n+1)∴a(n+1)/an=2当n=1时a2=a1=-2∴an={-2 n=1{-2^(n-1) n>=2∴sn=a1+a2+……+an=-2-(2+2^2+……+2^(n-1))=-2-[2(1-2^(n-1)/(1-2)]=-2^n