(1)a(n+2)-2a(n+1)+an=0a(n+2)-a(n+1)=a(n+1)-an,是等差数列d=(a4-a1)/(4-1)=-2an=10-2n(2)n<5时Sn=(8+10-2n)*n/2=(9-n)nn>=5时,Sn=-(8+10-2n)*n/2+2*(8+0)*5/2=-(9-n)n+40(3)存在虽然从某项开始Tn递减,但那是个收敛级数,存在最小值不过你题目有点问题,bn=1/(12-n)n,n取不到12……