2010年宝马525Li(E60)座椅电路图.pdf

12014届高考数学(理)一轮复习3数列求和一、{an}首项与公比分别是复数i+2(i是虚数单位)的实部与虚部,则数列{an}的前10项的和为()-1C.-20D.-2i解析:该等比数列的首项是2,公比是1,:{an}的通项公式是an=1n+n+1,若前n项和为10,则项数n为():an=1n+n+1=n+1-n,∴Sn=2-1+3-2+4-3+…+10-9+…+n+1-n=n+1-1=10,解得n=:(n)=n2n为奇数,-n2n为偶数,且an=f(n)+f(n+1),则a1+a2+a3+…+a100等于().-:由题意,a1+a2+…+a100=12-22-22+32+32-42-42+52+…+992-1002-1002+1012=-(1+2)+(3+2)+…-(99+100)+(101+100)=:(x)=x2+bx的图像在点A(1,f(1))处的切线的斜率为3,数列{1fn}的前n项和为Sn,则S2010的值为():∵f′(x)=2x+b,∴f′(1)=2+b=3,∴b=1,∴f(x)=x2+x,2∴1fn=1nn+1=1n-1n+1,∴S2010=1-12+12-13+…+12010-12011=1-12011=:{an}中,已知对任意正整数n,a1+a2+a3+…+an=2n-1,则a21+a22+a23+…+a2n等于()A.(2n-1)(2n-1)(4n-1)-1解析:∵a1+a2+a3+…+an=2n-1,∴a1+a2+…+an-1=2n-1-1,∴an=2n-2n-1=2n-1,∴a2n=4n-1,∴a21+a22+…+a2n=1-4n1-4=13(4n-1).答案:=logn+1(n+2)(n∈N*),若称使乘积a1·a2·a3·…·an为整数的数n为劣数,则在区间(1,2002)内所有的劣数的和为():设a1·a2·a3·…·an=lg3lg2·lg4lg3·…·lgn+2lgn+1=lgn+2lg2=log2(n+2)=k,则n=2k-2(k∈Z).令1<2k-2<2002,得k=2,3,4,…,10.∴所有劣数的和为41-291-2-18=211-22=:A二、+3+5+…+2x-111×2+12×3+…+1xx+1=110(x∈N*),则x=:原等式左边=x2x-1+1211-12+12-13+…+1x-1x+1=x2xx+1=x2+x=110,又x∈N*,所以x=:,1+2,1+2+22,1+2+22+23,…,1+2+22+23+…+2n-1,…:由题意得an=1+2+22+…+2n-1=1-2n1-2=2n-1,∴Sn=(21-1)+(22-1)+(23-1)+…+(2n-1)=(21+22+…+2n)-n=2-2n+11-2-n=2n+1-n-:2n+1-n-{an},定义数列{an+1-an}为数列{an}的“差数列”,若a1=2,{an}的“差数列”的通项公式为2n,则数列{an}的前n项和Sn=:∵an+1-an=2n,∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=2n-1+2n-2+…+22+2+2=2-2n1-2+2=2n-2+2=2n,∴Sn=2-2n+11-2=2n+1-:2n+1-、{an}满足a2=0,a6+a8=-10.(1)求数列{an}的通项公式;(2)求数列{an2n-1}:(1)设等差数列{an}的公差为d,由已知条件可得a1+d=0,2a1+12d=-10,解得a1=1,d=-{an}的通项公式为an=2-n.(2)设数列{an2n-1}的前n项和为Sn,即Sn=a1+a22+…+an2n-1,4故S1=1,Sn2=a12+a24+…+an2n,所以,当n>1时,Sn2=a1+a2-a12+…+an-an-12n-1-an2n=1-(12+14+…+12n-1)-2-n2n=1-(1-12n-1)-2-n2n==n2n-,数列{an2n-1}的前n项和Sn=n2n-