第04节 数列求和
A基础巩固训练
1.在等差数列word/media/image1_1.png中,word/media/image2_1.png=word/media/image3_1.png,则数列word/media/image1_1.png的前11项和word/media/image4_1.png=( ).
A.24 B.48 C.66 D.132
【答案】D
2.【2018届河南省郑州市第一中学高三一轮】已知正项等比数列word/media/image6_1.png的前word/media/image7_1.png项和为word/media/image8_1.png,且word/media/image9_1.png,则word/media/image10_1.png的最小值为( )
A. 10 B. 15 C. 20 D. 25
【答案】C
【解析】由题意可得: word/media/image11_1.png,由word/media/image12_1.png可得word/media/image13_1.png,
由等比数列的性质可得: word/media/image14_1.png成等比数列,
则: word/media/image15_1.png,综上可得:
word/media/image16_1.png,
当且仅当word/media/image17_1.png时等号成立.
综上可得,则word/media/image18_1.png的最小值为20.
本题选择C选项.
3.【2018届南宁二中、柳州高中高三9月联考】已知数列2008,2009,1,-2008,…若这个数列从第二项起,每一项都等于它的前后两项之和,则这个数列的前2018项之和word/media/image19_1.png__________.
【答案】4017
【解析】由题意可知word/media/image20_1.png
word/media/image21_1.png
所以word/media/image22_1.png即数列word/media/image23_1.png是以6为周期的数列,又word/media/image24_1.png
word/media/image25_1.png
4.【2017届浙江省丽水市高三下质量水平测试】已知数列word/media/image26_1.png的相邻两项word/media/image27_1.png是关于word/media/image28_1.png的方程word/media/image29_1.png的两实根,且word/media/image30_1.png.
(1)求word/media/image31_1.png的值;
(2)求证:数列word/media/image32_1.png是等比数列,并求数列word/media/image33_1.png的通项公式.
【答案】(1)word/media/image34_1.png, word/media/image35_1.png, word/media/image36_1.png (2)证明见解析, word/media/image37_1.png
试题解析:
(1)解:∵word/media/image39_1.png是关于word/media/image40_1.png的方程word/media/image41_1.png的两实根,
∴word/media/image42_1.png,因为word/media/image43_1.png,所以word/media/image44_1.png, word/media/image45_1.png, word/media/image46_1.png.
(2)∵word/media/image47_1.png,
故数列word/media/image48_1.png是首项为word/media/image49_1.png,公比为-1的等比数列.
所以word/media/image50_1.png,即word/media/image51_1.png.
5.【2018届安徽省蚌埠市第二中学高三7月月考】已知数列满足,.
(1)求数列的通项公式;
(2)证明:.
【答案】(1);(2)证明过程见解析
试题解析:(1)∵.
∴,∴是以为首项,2为公比的等比数列.
∴,即.
(2)证明:∵,,
∴.
B能力提升训练
1.已知两个等差数列word/media/image66_1.png和word/media/image67_1.png的前word/media/image68_1.png项和分别为word/media/image69_1.png和word/media/image70_1.png,且word/media/image71_1.png,则使得word/media/image72_1.png为整数的正整数word/media/image73_1.png的个数是( )
A.word/media/image74_1.png B.word/media/image75_1.png C.word/media/image76_1.png D.word/media/image77_1.png
【答案】C
【解析】由等差数列前word/media/image78_1.png项和的性质知,word/media/image79_1.png,故当word/media/image80_1.png,word/media/image81_1.png,word/media/image82_1.png,word/media/image83_1.png,word/media/image84_1.png时,word/media/image85_1.png为整数,故使得word/media/image86_1.png为整数的正整数word/media/image87_1.png的个数是word/media/image88_1.png.故应选C.
2.【2018届河南省洛阳市高三上尖子生第一次联考】已知数列word/media/image89_1.png满足word/media/image90_1.png,其中word/media/image91_1.png,若word/media/image92_1.png对word/media/image93_1.png恒成立,则实数word/media/image94_1.png的取值范围为__________.
【答案】word/media/image95_1.png
word/media/image97_1.png恒成立,即word/media/image98_1.png恒成立,当word/media/image99_1.png时, word/media/image100_1.png,而word/media/image101_1.png时word/media/image102_1.png,所以word/media/image103_1.png即可,当word/media/image104_1.png时, word/media/image105_1.png恒成立,综上word/media/image106_1.png,故填word/media/image107_1.png.
3.【2018届湖北省襄阳四中高三8月月考】用word/media/image108_1.png表示自然数word/media/image109_1.png的所有因数中最大的那个奇数,例如:9的因数有1,3,9,则word/media/image110_1.png的因数有1,2,5,10, word/media/image111_1.png,那么word/media/image112_1.png__________.
【答案】word/media/image113_1.png
【解析】由g(n)的定义易知g(n)=g(2n),且若n为奇数则g(n)=n
令f(n)=g(1)+g(2)+g(3)+…+g(2n-1)
则f(n+1)=g(1)+g(2)+g(3)+…g(2n+1-1)
=1+3+…+(2n+1-1)+g(2)+g(4)+…+g(2n+1-2)
=word/media/image114_1.png+g(1)+g(2)+…+g(2n+1-2)
=4n+f(n)
即f(n+1)-f(n)=4n,据此可得:
word/media/image115_1.png,
以上各式相加可得:
word/media/image116_1.png.
4.已知word/media/image117_1.png为数列word/media/image118_1.png前word/media/image119_1.png项和, 若word/media/image120_1.png,且word/media/image121_1.png,则word/media/image122_1.png .
【答案】5
5.已知数列word/media/image124_1.png的前word/media/image125_1.png项和为word/media/image126_1.png,点word/media/image127_1.png在抛物线word/media/image128_1.png上,各项都为正数的等比数列word/media/image129_1.png满足word/media/image130_1.png.
(Ⅰ)求数列word/media/image131_1.png, word/media/image132_1.png的通项公式;
(Ⅱ)记word/media/image133_1.png,求数列word/media/image134_1.png的前n项和word/media/image135_1.png.
【答案】(1)word/media/image136_1.png word/media/image137_1.png(2)word/media/image138_1.png
【解析】(Ⅰ)word/media/image139_1.png,当word/media/image140_1.png时, word/media/image141_1.png
当word/media/image142_1.png时, word/media/image143_1.png
word/media/image144_1.png, word/media/image145_1.png数列word/media/image146_1.png是首项为word/media/image147_1.png,公差为word/media/image148_1.png的等差数列, word/media/image149_1.png
又word/media/image150_1.png各项都为正数,解得word/media/image151_1.png
(Ⅱ)word/media/image152_1.png
C 思维拓展训练
1.已知函数word/media/image153_1.png的图像在点A(l,f(1))处的切线l与直线x十3y+2=0垂直,若数列word/media/image154_1.png的前n项和为word/media/image155_1.png,则word/media/image156_1.png的值为 ( )
A. word/media/image157_1.png B. word/media/image158_1.png C. word/media/image159_1.png D. word/media/image160_1.png
【答案】D
2.【2018届四川省成都七中高三上入学考试】设等差数列word/media/image162_1.png的前word/media/image163_1.png项和为word/media/image164_1.png,且word/media/image165_1.png(word/media/image166_1.png是常数, word/media/image167_1.png),word/media/image168_1.png,又word/media/image169_1.png,数列word/media/image170_1.png的前word/media/image171_1.png项和为word/media/image172_1.png,若word/media/image173_1.png对word/media/image174_1.png恒成立,则正整数word/media/image175_1.png的最大值是__________.
【答案】2
【解析】∵word/media/image176_1.png,
当n=1时, word/media/image177_1.png,
解得a1=2c,
当n=2时,S2=a2+a2−c,
即a1+a2=a2+a2−c,
解得a2=3c,∴3c=6,
解得c=2.
则a1=4,数列{an}的公差d=a2−a1=2,
∴an=a1+(n−1)d=2n+2.
∵word/media/image178_1.png
错位相减可得: word/media/image179_1.png,
则word/media/image180_1.png
∴数列{Tn}单调递增,T1最小,最小值为word/media/image181_1.png,
∴word/media/image182_1.png,
∴m<3,
故正整数m的最大值为2.
3.【2017届云南省昆明市高三下第二次统测】在平面直角坐标系上,有一点列word/media/image183_1.png,设点word/media/image184_1.png的坐标word/media/image185_1.png,其中word/media/image186_1.png,过点word/media/image187_1.png的直线与两坐标轴所围成的三角形面积为word/media/image188_1.png,设word/media/image189_1.png表示数列word/media/image190_1.png的前word/media/image191_1.png项和,则word/media/image192_1.png__________.
【答案】word/media/image193_1.png
4.【2017届广西南宁市金伦中学高三上期末】已知各项均为正数的数列word/media/image195_1.png的的前word/media/image196_1.png项和为word/media/image197_1.png,对word/media/image198_1.png,有word/media/image199_1.png.
(Ⅰ)求数列word/media/image200_1.png的通项公式;
(Ⅱ)令word/media/image201_1.png,设word/media/image202_1.png的前word/media/image203_1.png项和为word/media/image204_1.png,求证: word/media/image205_1.png.
【答案】(I)word/media/image206_1.png;(Ⅱ)证明过程见解析;
当word/media/image208_1.png时, word/media/image209_1.png, word/media/image210_1.png,两式相减得
word/media/image211_1.png,
所以数列word/media/image212_1.png是以1为首相,1为公差的等差数列, word/media/image213_1.png.
(Ⅱ)word/media/image214_1.png
word/media/image215_1.png
word/media/image216_1.png
word/media/image217_1.png
5.【2018届广东省揭阳市惠来县第一中学高三上第一次考试】记word/media/image218_1.png为差数列word/media/image219_1.png的前n项和,已知word/media/image220_1.png, word/media/image221_1.png.
(1)求word/media/image222_1.png的通项公式;
(2)令word/media/image223_1.png, word/media/image224_1.png,若word/media/image225_1.png对一切word/media/image226_1.png成立,求实数word/media/image227_1.png的最大值.
【答案】(1)word/media/image228_1.png(2)2
【解析】试题分析:(1)根据等差数列通项公式将条件转化为关于首项与公差的方程组,解方程组可得首项与公差的值,再代入通项公式即得word/media/image229_1.png的通项公式;(2)因为word/media/image230_1.png,所以利用裂项相消法可得word/media/image231_1.png,再根据word/media/image232_1.png最小值得word/media/image233_1.png,即得实数word/media/image234_1.png的最大值.
试题解析:解:(1)∵等差数列word/media/image235_1.png中, word/media/image236_1.png, word/media/image237_1.png.
∴word/media/image238_1.png,解得word/media/image239_1.png.
word/media/image240_1.png,
word/media/image241_1.png.
(2) word/media/image242_1.png,
word/media/image243_1.png,
word/media/image244_1.png随着word/media/image245_1.png增大而增大,
word/media/image246_1.png是递增数列, word/media/image247_1.png,
word/media/image248_1.png,
∴实数word/media/image249_1.png的最大值为2.
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