高二第一次教学检测答案(数学)
1--6.A C B C A B 7--12 .B D C D C D
13.3 14.60︒ 15.12 16.36
17.【解答】解:(1)∵△ABC中,cosA=>0,
∴A为锐角,sinA= = …
根据正弦定理,得 ,
∴ ,…
∴ …
(2)根据余弦定理,得a2=b2+c2﹣2bccosA,
∴9=4+c2﹣2×2c× ,
∴3c2﹣4c﹣15=0…
解之得:c=3或c=﹣ (舍去),
∴c=3
18解:(1)设{an}是公差为d的等差数列,
{bn}是公比为q的等比数列,
由b2=3,b3=9,可得q= =3,
bn=b2qn﹣2=3•3n﹣2=3n﹣1;
即有a1=b1=1,a14=b4=27,
则d= =2,
则an=a1+(n﹣1)d=1+2(n﹣1)=2n﹣1;
(2)cn=an+bn=2n﹣1+3n﹣1,
则数列{cn}的前n项和为
(1+3+…+(2n﹣1))+(1+3+9+…+3n﹣1)= n•2n+
=n2+
19.解:(1)由正弦定理可得:(2sinA+sinC)cosB=﹣sinBcosC,
∴2sinAcosB=﹣sinBcosC﹣cosBsinC=﹣sin(B+C)=﹣sinA,
又∵sinA>0,∴ ,
∵B∈(0,π),
∴
(2)b2=49=a2+c2﹣2accosB=a2+c2+ac=(a+c)2﹣ac=64﹣ac,
∴ac=15,
又∵a+c=8,∴
20.解:(1)利用正弦定理化简已知的等式得:2(a2﹣c2)=2 b(a﹣b),
整理得:a2﹣c2=ab﹣b2,即a2+b2﹣c2=ab,
∵c2=a2+b2﹣2abcosC,即a2+b2﹣c2=2abcosC,
∴2abcosC=ab,即cosC=,
则C=;
(2)∵C=,∴A+B=,即B=﹣A,
∵ = =2 ,即a=2 sinA,b=2 sinB,
∴S△ABC= absinC= absin = ×2 sinA×2 sinB×
=2 sinAsinB=2 sinAsin( ﹣A)=2 sinA( cosA+ sinA)
=3sinAcosA+ sin2A= sin2A+ (1﹣cos2A)
= sin2A﹣ cos2A+ = sin(2A﹣ )+,
则当2A﹣ =,即A=时,S△ABCmax=.
21.解:(1)证明:t=0,m=0时,an=2an﹣1+2n,
两边同除以2n,可得 = +1,
即有 是首项为 ,公差为1的等差数列;
(2)证明:t=﹣1,m=时,an=2an﹣1+3,
两边同加上3,可得an+3=2(an﹣1+3),
即有数列{an+3}为首项为6,公比为2的等比数列;
(3)t=0,m=1时,an=2an﹣1+2n+3,
两边同除以2n,可得 = +1+
即为 = =1+ ,
即有得 = +( ﹣ )+( ﹣ )+…+( ﹣ )
= +1+ +1+ +…+1+ ,
=n﹣1+ =n+2﹣ ,
则an=(n+2)•2n﹣3,
前n项和Sn=3•2+4•22+5•23+…+(n+2)•2n﹣3n,
可令Rn=3•2+4•22+5•23+…+(n+2)•2n,
2Rn=3•22+4•23+5•24+…+(n+2)•2n+1,
两式相减可得,﹣Rn=3•2+22+23+…+2n﹣(n+2)•2n+1
=4+ ﹣(n+2)•2n+1
=2﹣(n+1)•2n+1,
则Rn═(n+1)•2n+1﹣2,
Sn=(n+1)•2n+1﹣2﹣3n.
22.解:(1)当n=1时, ,
∵
(2)当n≥2时,满足 ,且 ,
∴ ,
∴ ,
∵an>0,∴an+1=an+2,
∴当n≥2时,{an}是公差d=2的等差数列.
∵a2,a5,a14构成等比数列,∴ , ,解得a2=3,
由(1)可知, ,∴a1=1∵a2﹣a1=3﹣1=2,
∴{an}是首项a1=1,公差d=2的等差数列.
∴数列{an}的通项公式an=2n﹣1.
(3)由(2)可得式 =.
∴
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