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1、考点规范练31 数列求和 考点规范练A册第20页 基础巩固组1.数列112,314,518,7116,…,(2n-1)+12n,…的前n项和Sn的值等于( ) A.n2+1-12nB.2n2-n+1-12nC.n2+1-12n-1D.n2-n+1-12n答案:A解析:该数列的通项公式为an=(2n-1)+12n,则Sn=[1+3+5+…+(2n-1)]+12+122+…+12n=n2+1-12n.2.(2015云南曲靖一模)122-1+132-1+142-1+…+1(n+1)2-1的值为( )A.n+12(n+2)B.34-n+12(n+2)C
2、.34-121n+1+1n+2D.32-1n+1+1n+2〚导学号92950494〛答案:C解析:∵1(n+1)2-1=1n2+2n=1n(n+2)=121n-1n+2,∴122-1+132-1+142-1+…+1(n+1)2-1=121-13+12-14+13-15+…+1n-1n+2=1232-1n+1-1n+2=34-121n+1+1n+2.3.已知数列{an}:12,13+23,14+24+34,…,110+210+310+…+910,…,若bn=1anan+1,那么数列{bn}的前n项和Sn等于( )A.nn+1B.4nn+1C.3nn+1D.5nn+1答案:B解析:易得an
3、=1+2+3+…+nn+1=n2,∴bn=1anan+1=4n(n+1)=41n-1n+1.∴Sn=41-12+12-13+…+1n-1n+1=41-1n+1=4nn+1.4.已知函数f(x)=xa的图像过点(4,2),令an=1f(n+1)+f(n),n∈N+.记数列{an}的前n项和为Sn,则S2016等于( )A.2016-1B.2016+1C.2017-1D.2017+1答案:C解析:由f(4)=2,可得4a=2,解得a=12,则f(x)=x12.∴an=1f(n+1)+f(n)=1n+1+n=n+1-n,S2016=a1+a2+a3+…+a2016=(2-1)+(3-2)+(
4、4-3)+…+(2017-2016)=2017-1.5.数列{an}满足an+1+(-1)nan=2n-1,则{an}的前60项和为( )A.3690B.3660C.1845D.1830答案:D解析:∵an+1+(-1)nan=2n-1,∴当n=2k(k∈N+)时,a2k+1+a2k=4k-1,①当n=2k+1(k∈N)时,a2k+2-a2k+1=4k+1,②4①+②得:a2k+a2k+2=8k.则a2+a4+a6+a8+…+a60=(a2+a4)+(a6+a8)+…+(a58+a60)=8(1+3+…+29)=8×15×(1+29)2=1800.由②得a2k+1=a2k+2-(4k+
5、1),∴a1+a3+a5+…+a59=a2+a4+…+a60-[4×(0+1+2+…+29)+30]=1800-4×30×292+30=30,∴a1+a2+…+a60=1800+30=1830.6.3·2-1+4·2-2+5·2-3+…+(n+2)·2-n= .〚导学号92950495〛 答案:4-n+42n解析:设Sn=3·2-1+4·2-2+5·2-3+…+(n+2)·2-n,①则12Sn=3·2-2+4·2-3+…+(n+1)2-n+(n+2)2-n-1.②①-②,得12Sn=3·2-1+2-2+2-3+…+2-n-(n+2)·2-n-1=2·2-1+2-1+2-2+2-3
6、+…+2-n-(n+2)·2-n-1=1+2-1(1-2-n)1-2-1-(n+2)·2-n-1=2-(n+4)·2-n-1.故Sn=4-n+42n.7.已知在等比数列{an}中,a1=3,a4=81,若数列{bn}满足bn=log3an,则数列1bnbn+1的前n项和Sn= . 答案:nn+1解析:设等比数列{an}的公比为q,则a4a1=q3=27,解得q=3.∴an=a1qn-1=3×3n-1=3n,故bn=log3an=n,∴1bnbn+1=1n(n+1)=1n-1n+1,则数列1bnbn+1的前n项和为1-12+12-13+…+1n-1n+1=1-1n+1=nn+1.8
7、.(2015长春模拟)已知等差数列{an}满足:a5=9,a2+a6=14.(1)求{an}的通项公式;(2)若bn=an+qan(q>0),求数列{bn}的前n项和Sn.解:(1)∵在等差数列{an}中,a5=9,a2+a6=2a4=14,∴a4=7,其公差d=a5-a4=2,∴an=a4+(n-4)d=7+2(n-4)=2n-1.(2)∵bn=an+qan(q>0),∴Sn=b1+b2+…+bn=(a1+a2+…+an)+(qa
