Simone
（I)Sn=(an+an^2)/2S(n-1)=[a(n-1)+a(n-1)^2]/2an=Sn-S(n-1)={[an-a(n-1)]+[an^2-a(n-1)^2]}/22an=an-a(n-1)+an^2-a(n-1)^2an+a(n-1)=[an+a(n-1)][an-a(n-1)]由于各项都是正数，因此an+a(n-1)>0,两边同时除以该式世哪判搜改，得an-a(n-1)=1等差数列，公差为1，a1=1,an=n验证：Sn=n(n+1)/缓备2=(n^2+n)/2=(an^2+an)/2(II)bn=an/2^n=n/2^nTn=1/2+2/4+3/8+4/16+...+n/2^nTn/2=1/4+2/8+3/16+...+n/2^(n+1)Tn-Tn/2=1/2+1/4+1/8+1/16+...+1/2^n-n/2^(n+1)=(1/2)(1-1/2^n)/(1-1/2)-n/2^(n+1)=1-1/2^n-n/2^(n+1)0
