2018_2019版高中數(shù)學(xué)第二章數(shù)列2.3.1等差數(shù)列的前n項(xiàng)和練習(xí)新人教A版必修.docx

第1課時(shí)等差數(shù)列的前n項(xiàng)和課后篇鞏固探究A組1.若等差數(shù)列an的前5項(xiàng)和S5=25,且a2=3,則a7=()A.12B.13C.14D.15解析S5=25,a2+a4=10.又a2=3,a4=7,公差d=2.a7=a4+3d=7+32=13.答案B2.在等差數(shù)列an中,已知a4+a8=16,則該數(shù)列前11項(xiàng)的和S11=()A.58B.88C.143D.176解析S11=,a1+a11=a4+a8=16,S11=88,故選B.答案B3.設(shè)Sn為等差數(shù)列an的前n項(xiàng)和,若a1=1,公差d=2,Sk+2-Sk=24,則k=()A.8B.7C.6D.5解析由a1=1,公差d=2,得an=2n-1.又Sk+2-Sk=ak+1+ak+2,所以2k+1+2k+3=24,得k=5,故選D.答案D4.若公差不為0的等差數(shù)列an的前21項(xiàng)的和等于前8項(xiàng)的和,且a8+ak=0,則正整數(shù)k的值為()A.20B.21C.22D.23解析設(shè)等差數(shù)列an的前n項(xiàng)和為Sn,由題意,得S21=S8,即a9+a10+a21=0.根據(jù)等差數(shù)列的性質(zhì),得13a15=0,即a15=0.故a8+a22=2a15=0,即k=22.故選C.答案C5.已知數(shù)列an的通項(xiàng)公式為an=2n+1,令bn= (a1+a2+an),則數(shù)列bn的前10項(xiàng)和T10=()A.70B.75C.80D.85解析an=2n+1,數(shù)列an是等差數(shù)列,首項(xiàng)a1=3,其前n項(xiàng)和Sn=n2+2n,bn=Sn=n+2,數(shù)列bn也是等差數(shù)列,首項(xiàng)b1=3,公差為1,其前10項(xiàng)和T10=103+1=75,故選B.答案B6.設(shè)數(shù)列an是等差數(shù)列,且a2+a3+a4=15,則該數(shù)列的前5項(xiàng)和S5=.解析由a2+a3+a4=15,得3a3=15,解得a3=5,故S5=5a3=25.答案257.在等差數(shù)列an中,其前n項(xiàng)和為Sn,若S12=8S4,則=.解析S12=12a1+d,S4=4a1+d,12a1+66d=32a1+48d.20a1=18d.答案8.已知數(shù)列an的前n項(xiàng)和為Sn=n2n-1,則a3+a4+a5=.解析a3+a4+a5=S5-S2=(525-1)-(222-1)=152.答案1529.導(dǎo)學(xué)號(hào)04994034設(shè)數(shù)列an的前n項(xiàng)和為Sn,點(diǎn)(nN*)均在函數(shù)y=3x-2的圖象上,求數(shù)列an的通項(xiàng)公式.解依題意,得=3n-2,即Sn=3n2-2n.當(dāng)n2時(shí),an=Sn-Sn-1=(3n2-2n)-3(n-1)2-2(n-1)=6n-5.因?yàn)閍1=S1=1,滿足an=6n-5,所以an=6n-5(nN*).10.(2017江西上高二中期末)已知數(shù)列an滿足a1=1,a2=2,an+2=2an+1-an+2.(1)設(shè)bn=an+1-an,證明bn是等差數(shù)列;(2)求an的通項(xiàng)公式.解(1)an+2=2an+1-an+2,an+2-an+1=an+1-an+2,即bn+1=bn+2.又b1=a2-a1=2-1=1,數(shù)列bn是以1為首項(xiàng),2為公差的等差數(shù)列.(2)由(1)可知,an+1-an=1+2(n-1)=2n-1,an-an-1=2(n-1)-1,an-1-an-2=2(n-2)-1,a2-a1=21-1,累加,得an-a1=2-(n-1)=n2-2n+1,an=a1+n2-2n+1=n2-2n+2,數(shù)列an的通項(xiàng)公式為an=n2-2n+2.B組1.在等差數(shù)列an中,2a4+a7=3,則數(shù)列an的前9項(xiàng)和S9等于()A.3B.6C.9D.12解析設(shè)等差數(shù)列an的公差為d,因?yàn)?a4+a7=3,所以2(a1+3d)+a1+6d=3,整理,得a1+4d=1,即a5=1,所以S9=9a5=9.答案C2.已知數(shù)列an的前n項(xiàng)和Sn=n2,則an等于()A.nB.n2C.2n+1D.2n-1解析當(dāng)n=1時(shí),a1=S1=1;當(dāng)n2時(shí),an=Sn-Sn-1=n2-(n-1)2=2n-1,且a1=1適合上式,故an=2n-1(nN*).答案D3.已知等差數(shù)列an,a2=6,a5=15,若bn=a2n,則數(shù)列bn的前5項(xiàng)和等于()A.30B.45C.90D.186解析由等差數(shù)列an易得公差d1=3.又bn=a2n,所以bn也是等差數(shù)列,公差d2=6.故S5=b1+b2+b3+b4+b5=a2+a4+a6+a8+a10=56+6=90.答案C4.設(shè)Sn為等差數(shù)列an的前n項(xiàng)和,Sn=336,a2+a5+a8=6,an-4=30(n5,nN*),則n等于()A.8B.16C.21D.32解析由a2+a5+a8=6,得3a5=6,所以a5=2.因?yàn)閍5+an-4=a1+an=2+30=32,所以Sn=336,解得n=21.答案C5.已知數(shù)列an的前n項(xiàng)和為Sn,且Sn=2an-1(nN*),則a5=.解析當(dāng)n2時(shí),由Sn=2an-1,得Sn-1=2an-1-1.兩式相減,得an=2an-2an-1,所以an=2an-1.因?yàn)閍1=2a1-1,所以a1=1,故a5=2a4=22a3=23a2=24a1=16.答案166.在數(shù)列an中,an=4n-,a1+a2+an=an2+bn+c,nN*,其中a,b為常數(shù),則ab+c=.解析因?yàn)閍n=4n-,即an是關(guān)于n的一次函數(shù),所以數(shù)列an是等差數(shù)列,所以a1+a2+an=2n2-n,因此a=2,b=-,c=0,故ab+c=2+0=-1.答案-17.已知數(shù)列an的前n項(xiàng)和為Sn(Sn0),且滿足an+2SnSn-1=0(n2),a1=.(1)求證:是等差數(shù)列;(2)求數(shù)列an的通項(xiàng)公式.(1)證明-an=2SnSn-1(n2),-Sn+Sn-1=2SnSn-1(n2).又Sn0(n=1,2,3,),=2.又=2,是以2為首項(xiàng),2為公差的等差數(shù)列.(2)解由(1)可知=2+(n-1)2=2n,Sn=.當(dāng)n2時(shí),an=Sn-Sn-1=-;當(dāng)n=1時(shí),S1=a1=.故an=8.導(dǎo)學(xué)號(hào)04994035設(shè)Sn為數(shù)列an的前n項(xiàng)和,Sn=an-1(為常數(shù),n=1,2,3,).(1)若a3=,求的值;(2)是否存在實(shí)數(shù),使得數(shù)列an是等差數(shù)列?若存在,求出的值;若不存在,請(qǐng)說(shuō)明理由.解(1)因?yàn)镾n=an-1,所以a1=a1-1,a2+a1=a