2023山东济宁高三一模答案汇总-生物答案
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18.解析:(1)由题意知Sn=n2+2n.当n≥2时,an=Sn-Sn-1=2n+1;(3分)当n=1时,a1=S1=3,适合上式所以an=2n+1.(5分)(2)bn=2anan+1(2n+1)(2n+3)2n+12n+3(8分)则b+b2+…+b2=111_1+…1116221321217(12分)
8.B【解析】∵点P(3,0)为圆C:(x-3)2+y2=4的圆,PA·PB=2,∴PA|XPB|Xcos∠APB=2×cos∠APB=2,∴cos∠APB),又∵0≤∠APB≤180°,∴∠APB=60°,∴△APB为等边三角形,∴直线l被圆C截得线段的长是2.故选B.
17.【命题意图】木题考查等差、等比数列的性质,数列的前n项秈【解析】()设数列an}的公比为q(q≠0).因为S是3a2和a4的等差中项,所以25=3a2+a,…(1分)即2(a1+a1q+a1q2)=3a1q+a1q3将理得2(1+q2)=q(1+q2),解得q=2.……………(3分)所以m1…(4分)所以an=2×2°=2(6分)(Ⅱ)由(1)可知b。=(7分)所以7+T2+2,¨…………(9分)①-②,可=×上4+、1?22=1-2+2所以7。=2(12分)
18解:(1)由A(4,3)及AB边上的高所在直线方程为得AB所在直线方程为x+y-7=0.(3分)又BD所在直线方程为3x+y-7=0由得B(0,7分)+y-7=0(2)设C(m,n),又A(4,3),D为AC中点,则Dm+4n+33m+4n+37=0由已知得3=0得C(2,一2)(10分)又B(0,7),所以BC边所在的直线方程为19x+y-7=0(12分)
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