《高三理科數(shù)學(xué) 二輪復(fù)習(xí)跟蹤強(qiáng)化訓(xùn)練:19 Word版含解析》由會員分享,可在線閱讀,更多相關(guān)《高三理科數(shù)學(xué) 二輪復(fù)習(xí)跟蹤強(qiáng)化訓(xùn)練:19 Word版含解析(4頁珍藏版)》請?jiān)谘b配圖網(wǎng)上搜索。
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跟蹤強(qiáng)化訓(xùn)練(十九)
1.(20xx沈陽質(zhì)檢)已知數(shù)列{an}是公差不為0的等差數(shù)列,首項(xiàng)a1=1,且a1,a2,a4成等比數(shù)列.
(1)求數(shù)列{an}的通項(xiàng)公式;
(2)設(shè)數(shù)列{bn}滿足bn=an+,求數(shù)列{bn}的前n項(xiàng)和Tn.
[解] (1)設(shè)數(shù)列{an}的公差為d,由已知得,a=a1a4,
即(1+d)2=1+3d,解得d=0或d=1.
又d≠0,∴d=1,可得an=n.
(2)由(1)得bn=n+2n,
∴Tn=(1+21)+(2+22)+(3+23)+…+(n+2n)
=(1+2+3+…+n)+(2+22+23+…
2、+2n)
=+2n+1-2.
[解] (1)由題意得,解得
當(dāng)n≥2時,Sn-1=(n-1)an-(n-1)n,
所以an=nan+1-n(n+1)-(n-1)an+(n-1)n,
即an+1-an=2.
又a2-a1=2,因而數(shù)列{an}是首項(xiàng)為1,公差為2的等差數(shù)列,
從而an=2n-1.
Tn=121+322+523+…+(2n-3)2n-1+(2n-1)2n,
2Tn=122+323+524+…+(2n-3)2n+(2n-1)2n+1.
兩式相減得
-Tn=121+222+223+…+22n-(2n-1)2n+1
=-2+2(21+22+23+…+2n
3、)-(2n-1)2n+1
=-2+2-(2n-1)2n+1
=-2+2n+2-4-(2n-1)2n+1=-6-(2n-3)2n+1.
所以Tn=6+(2n-3)2n+1.
3.?dāng)?shù)列{an}的前n項(xiàng)和為Sn,且首項(xiàng)a1≠3,an+1=Sn+3n(n∈N*).
(1)求證:{Sn-3n}是等比數(shù)列;
(2)若{an}為遞增數(shù)列,求a1的取值范圍.
[解] (1)證明:∵an+1=Sn+3n,(n∈N*)
∴Sn+1=2Sn+3n,
∴Sn+1-3n+1=2(Sn-3n),∵a1≠3.
∴=2,
∴數(shù)列{Sn-3n}是公比為2,首項(xiàng)為a1-3的等比數(shù)列.
(2)由(1)得Sn
4、-3n=(a1-3)2n-1,∴Sn=(a1-3)2n-1+3n,
∴當(dāng)n≥2時,an=Sn-Sn-1=(a1-3)2n-2+23n-1,∵{an}為遞增數(shù)列,
∴n≥2時,(a1-3)2n-1+23n>(a1-3)2n-2+23n-1,∴n≥2時,
2n-2>0,
可得n≥2時,a1>3-12n-2,
又當(dāng)n=2時,3-12n-2有最大值為-9,
∴a1>-9,又a2=a1+3滿足a2>a1,
∴a1的取值范圍是(-9,+∞).
4.(20xx昆明模擬)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,a1=1,當(dāng)n≥2時,an=2anSn-2S.
(1)求數(shù)列{an}的通項(xiàng)公式;
(2)
5、是否存在正數(shù)k,使(1+S1)(1+S2)…(1+Sn)≥k對一切正整數(shù)n都成立?若存在,求k的取值范圍;若不存在,請說明理由.
[解] (1)∵當(dāng)n≥2時,an=Sn-Sn-1,an=2anSn-2S,
∴Sn-Sn-1=2(Sn-Sn-1)Sn-2S.
∴Sn-1-Sn=2SnSn-1.
∴-=2.
∴數(shù)列是首項(xiàng)為==1,公差為2的等差數(shù)列,
即=1+(n-1)2=2n-1.
∴Sn=.
當(dāng)n≥2時,an=Sn-Sn-1=-
=.
∴數(shù)列{an}的通項(xiàng)公式為an=
(2)設(shè)bn=,
則bn+1=.
由(1)知Sn=,Sn+1=,
∴==
= >1.
又bn>0,∴數(shù)列{bn}是單調(diào)遞增數(shù)列.
由(1+S1)(1+S2)…(1+Sn)≥k,得bn≥k.
∴k≤b1==.
∴存在正數(shù)k,使(1+S1)(1+S2)…(1+Sn)≥k對一切正整數(shù)n都成立,且k的取值范圍為.