Question

设数列{an}的前n项和为SN,点(N,SN/N)均在函数Y=-X+12的图像上

Answer 1

(1) Sn/n=-n+12=>Sn=-n²+12n

(2) an=Sn-S(n-1)=-n²+12n+(n-1)²-12(n-1)=-2n+1+12=-2n+13

所以an-a(n-1)=-2n+13+2(n-1)-13=-2,所以an是等差数列,公差为-2

(3) T16=|a1|+|a2|+..+|a16|=11+9+7+...+1+1+3+.+19=36+100=136

(4) bn=(an-13)/2=-n,Sn-3 -n²+12n-3 m+12