用特殊值：a1+S1=1，且a1=S1得a1=1/2
进一步求得an=(1/2)^n
bn=2+(1/2)^n
an=(1/2)^n
bn=2+(1/2)^n
a1+S1=1.a1=S1
得a1=0.5
(an+Sn)-(a<n-1>+S<n-1>)=an-a<n-1>+Sn-S<n-1>
=2an-a<n-1>=1-1=0
得2an=a<n-1>即an=0.5a<n-1>=(0.5)^n:0.5的n次方
(2)
b(n+1)=bn+an=b(n-1)+an+a(n-1)=.....=b1+sn=b1+1-an
=2-an=2-(0.5)^n