1第一章数列习题课2数列的求和问题课后篇巩固提升必备知识基础练1.已知数列{an}的通项an=2n+1,n∈N+,由bn=a1+a2+a3+…+ann所确定的数列{bn}的前n项的和是()A.n(n+2)B.12n(n+4)C.12n(n+5)D.12n(n+7)答案C解析 a1+a2+…+an=n2(2n+4)=n2+2n,∴bn=n+2,∴{bn}的前n项和Sn=n\(n+5\)2.2.数列12×5,15×8,18×11,…,1\(3n-1\)×\(3n+2\),…的前n项和为()A.n3n+2B.n6n+4C.3n6n+4D.n+1n+2答案B解析由数列通项公式1\(3n-1\)\(3n+2\)=1313n-1−13n+2,得前n项和Sn=1312−15+15−18+18−111+…+13n-1−13n+2=1312−13n+2=n6n+4.3.1+1+12+1+12+14+…+1+12+14+…+1210的值为()A.18+129B.20+1210C.22+1211D.18+1210答案B2解析设an=1+12+14+…+12n-1=1×[1-(12)n]1-12=21-12n,∴原式=a1+a2+…+a11=21-121+21-122+…+21-1211=211-12+122+…+1211=211-12(1-1211)1-12=210+1211=20+1210.4.设an=1√n+1+√n,数列{an}的前n项和Sn=9,则n=.答案99解析an=1√n+1+√n=√n+1−√n,故Sn=√2-1+√3−√2+…+√n+1−√n=√n+1-1=9.解得n=99.5.在数列{an}中,已知Sn=1-5+9-13+17-21+…+(-1)n-1(4n-3),n∈N+,则S15+S22-S31的值是.答案-76解析S15=-4×7+a15=-28+57=29,S22=-4×11=-44,S31=-4×15+a31=-60+121=61,S15+S22-S31=29-44-61=-76.6.已知函数f(x)=2x-3x-1,点(n,an)(n∈N+)在f(x)的图象上,数列{an}的前n项和为Sn,求Sn.解由题意得an=2n-3n-1,n∈N+,Sn=a1+a2+…+an=(2+22+…+2n)-3(1+2+3+…+n)-n=2\(1-2n\)1-2-3·n\(n+1\)2-n=2n+1-n\(3n+5\)2-2.7.已知等差数列{an}中,2a2+a3+a5=20,且前10项和S10=100.(1)求数列{an}的通项公式;(2)若bn=1anan+1,求数列{bn}的前n项和.解(1)设数列{an}的通项公式为an=a1+(n-1)d,3由已知得{2a2+a3+a5=4a1+8d=20,10a1+10×92d=10a1+45d=100,解得{a1=1,d=2,所以数列{an}的通项公式为an=1+2(n-1)=2n-1.(2)bn=1\(2n-1\)\(2n+1\)=1212n-1−12n+1,所以Tn=121-13+13−15+…+12n-1−12n+1=121-12n+1=n2n+1.关键能力提升练8.已知函数f(x)=21+x2(x∈R),若等比数列{an}满足a1a2021=1,则f(a1)+f(a2)+f(a3)+…+f(a2021)=()A.2021B.20212C.2D.12答案A解析 函数f(x)=21+x2(x∈R),∴f(x)+f1x=21+x2+21+(1x)2=21+x2+2x2x2+1=2. 数列{an}为等比数列,且a1·a2021=1,∴a1a2021=a2a2020=a3a2019=…=a2021a1=1.∴f(a1)+f(a2021)=f(a2)+f(a2020)=f(a3)+f(a2019)=…=f(a2021)+f(a1)=2,∴f(a1)+f(a2)+f(a3)+…+f(a2021)=2021.故选A.9.已知等差数列{an}中,a3+a5=a4+7,a10=19,...
