18.D [解析] 由题意可得an+1-an=n+1,则a1=1,a2-a1=2,a3-a2=3,…,an-an-1=n,以上各式相加可得
an=
,则=2
-,所以++…+
=2×1-
+
-+…+ -
= .故
选D.
19.D [解析] 当n≥2时,2Sn=2(Sn-1+an)=an+ ,即 +2Sn-1an-1=0,所以an=-Sn-1± .又由-
an>0得an=-Sn-1+ ,则S ,所以 - =1,即数列{ }是公差为1的等n=an+Sn-1= - - - 差数列,又2S1=2a1=a1+ ,解得a1=1(舍去a1=-1),即S1=1, =1,所以 =n,所以S2018= .故选
D.
20.[122,162] [解析] 设4个实数根依次为m,m+d,m+2d,m+3d,不妨设 m,m+3d为x-18x+a=0的两个实数根,则m+d,m+2d为方程x-18x+b=0的两个根,由韦达定理得2m+3d=18,即m=9-d,又
2
2
m(m+3d)=a,(m+d)(m+2d)=b,
故a+b=9-d
9+d+9-d
9+d=81-d+81-d=162-d,又因为d∈[0,16],所以a+b∈
2
2
2
2
[122,162],即a+b的取值范围是[122,162].