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1、2.2等差数列自主广场我夯基我达标1.已知数列{an}的通项公式为an=2(n+1)+3,则此数列()A.是公差为2的等差数列B.是公差为3的等差数列C.是公差为5的等差数列D.不是等差数列思路解析:已知a1=7,an-an-1=2(n≥2),故这是一个以2为公差的等差数列.也可根据等差数列通项公式an=dn+(a1-d)知d=2.答案:A2.在等差数列{an}中,已知a1=2,a2+a3=13,则a4+a5+a6等于()A.40B.42C.43D.45思路解析:在等差数列{an}中,已知a1=2,a2+a3=13,∴d=3.∴a5=14.则a4+a5+a6=3a5=4
2、2.答案:B3.设Sn是等差数列{an}的前n项和,若S7=35,则a4等于()A.8B.7C.6D.5思路解析:S7=7a4=35,∴a4=5.答案:D4.已知等差数列{an}中,a2=7,a4=15,则前10项的和S10等于()A.100B.210C.380D.400思路解析:d==4,a1=3,所以S10=210.答案:B5.等差数列{an}的前m项和为30,前2m项和为100,则它的前3m项和为()A.130B.170C.210D.260思路解析:令m=1,则Sm=S1=a1=30,S2m=S2=a1+a2=100,则有a1=30,a2=70,d=40,则a3=
3、110,故S3m=S3=S2+a3=100+110=210.注:也可以用Sm,S2m-Sm,S3m-S2m成等差数列求解.答案:C6.在各项均不为零的等差数列{an}中,若an+1-an2+an-1=0(n≥2),则S2n-1-4n等于()A.-2B.0C.1D.2思路解析:设公差为d,则an+1=an+d,an-1=an-d,由an+1-an2+an-1=0(n≥2),得2an-an2=0,解得an=2(零解舍去),故S2n-1-4n=2×(2n-1)-4n=-2.答案:A7.设Sn为等差数列{an}的前n项和,S4=14,S10-S7=30,则S9=________
4、___.思路解析:设等差数列{an}的首项为a1,公差为d,由题意,得4a1+=14,［10a1+］-［7a1+］=30,联立解得a1=2,d=1,所以S9=9×2+·1=54.答案:548.若两个等差数列的前n项和之比是(7n+1)∶(4n+27),则它们的第11项之比为_____________.思路解析:方法一:设数列{an}的前n项和为Sn,数列{bn}的前n项和为Tn,则a11=,b11=,∴.方法二:等差数列前n项和是关于n的二次函数,则可设Sn=(7n+1)·nk,Tn=(4n+27)·nk,由an=Sn-Sn-1=k(14n-6),得a11=148k,n
5、≥2.bn=Tn-Tn-1=k(8n+23),得b11=111k.∴.答案:我综合我发展9.设数列{an}、{bn}、{cn}满足:bn=an-an+2,cn=an+2an+1+3an+2(n=1,2,3,…).求证:{an}为等差数列的充分必要条件是{cn}为等差数列且bn≤bn+1(n=1,2,3,…).证明:必要性:设{an}是公差为d1的等差数列,则bn+1-bn=(an+1-an+3)-(an-an+2)=(an+1-an)-(an+3-an+2)=d1-d1=0.∴bn≤bn+1(n=1,2,3,…)成立.又cn+1-cn=(an+1-an)+2(an+2-
6、an+1)+3(an+3-an+2)=d1+2d1+3d1=6d1(常数)(n=1,2,3,…),∴数列{cn}为等差数列.充分性:设数列{cn}是公差为d2的等差数列,且bn≤bn+1(n=1,2,3,…),∵cn=an+2an+1+3an+2,①∴cn+2=an+2+2an+3+3an+4.②①-②,得cn-cn+2=(an-an+2)+2(an+1-an+3)+3(an+2-an+4)=bn+2bn+1+3bn+2.∵cn-cn+2=(cn-cn+1)+(cn+1-cn+2)=-2d2,∴bn+2bn+1+3bn+2=-2d2.③从而有bn+1+2bn+2+3bn
7、+3=-2d2.④④-③,得(bn+1-bn)+2(bn+2-bn+1)+3(bn+3-bn+2)=0.⑤∵bn+1-bn≥0,bn+2-bn+1≥0,bn+3-bn+2≥0,∴由⑤得bn+1-bn=0(n=1,2,3,…).由此不妨设bn=d3(n=1,2,3,…),则an-an+2=d3(常数).由此cn=an+2an+1+3an+2=4an+2an+1-3d3.⑥从而cn+1=4an+1+2an+2-3d3,⑦⑦-⑥,得cn+1-cn=2(an+1-an)-2d3.因此an+1-an=(cn+1-cn)+d3=d2+d3(常数)(n
