a(n+1)=sn
an=s(n-1)
两式相减得
a(n+1)-an=an
a(n+1)=2an
a(n+1)/an=2
所以an 是以2为公比的等数列
an=a1q^(n-1)
=-2*2^(n-1)
=-2^n
sn=a1(1-q^n)/(1-q)
=-2*(1-2^n)/(1-2)
=-2(2^n-1)
=2-2^(n+1)