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通券婷2242Sn为数列{an}的前n项和,已知an>0,an2+2an=4Sn+3.(1)求{an}的通项公式;(2)设bn=1anan+1,设数列{bn}前n项和Tn,且λ≤Tn对一切n∈N*都成立,试求... -
咎兰融18697313413 ______[答案] (1)由an2+2an=4Sn+3,① 可知an-12+2an-1=4Sn-1+3,②(n≥2) ①-②得:an2-an-12+2an-2an-1=4an, 即(an+an-1)... ∴{an}是以a1=3为首项,d=2为公差的等差数列. ∴an=2n+1(n∈N*). (2)bn= 1 anan+1= 1 (2n+1)(2n+3)= 1 2( 1 2n+1- 1 2n+3)...

通券婷2242已知数列{an}中,a1=1,an+1=2an/(an+2)求数列{anan+1}的前n项和Sn -
咎兰融18697313413 ______[答案] 由已知可得1/(an+1)=1/an+1/2 易得1/an=(n+1)/2 所以an=2/(n+1) anan+1=2(1/n-1/(n+1)) 故Sn=2n/(n+1)

通券婷2242An=3n - 2,求数列{1/AnAn+1}的前n项和n和n+1是下脚标,应该都明白吧 -
咎兰融18697313413 ______[答案] 考察一般项第k项: 1/[aka(k+1)]=1/[(3k-2)(3(k+1)-2)]=1/[(3k-2)(3k+1)]=(1/3)[1/(3k-2)-1/(3k+1)] 1/(a1a2)+1/(a2a3)+...+1/[ana(n+1)] =(1/3)[1/1-1/4+1/4-1/7+...+1/(3n-2)-1/(3n+1)] =(1/3)[1-1/(3n+1)] =n/(3n+1)

通券婷2242已知各项均为正数的数列an满足a1=1an+1+anan+1 - an=0 -
咎兰融18697313413 ______[答案] a(n+1)+ana(n+1)-an=0 1/an + 1- 1/a(n+1) =0 1/a(n+1) -1/an = 1 {1/an}是等差数列,d=1 1/an - 1/a1= n-1 1/an = n an = 1/n

通券婷2242已知各项均为正数的数列{an}的首项a1=1,Sn是数列{an}的前n项和,且满足:anSn+1 - an+1Sn+an - an+1= 1 2 anan+1,则 3 34 S12=___. -
咎兰融18697313413 ______[答案] ∵anSn+1-an+1Sn+an-an+1= 1 2anan+1,且Sn+1=Sn+an+1, ∴(an-an+1)Sn+ 1 2anan+1+an-an+1=0, ∴Sn+ anan+1 2(an-an+1)+1=0; 又∵a1=1,令n=1,则1+ a2 2(1-a2)+1=0,解得a2= 4 3, 同理可得a3= 5 3, 猜想an= n+2 3; 下面利用数学...

通券婷2242大侠 证明能被7整除的特征若7|(AnAn - 1…A6A5A4 - A3A2A1),则7|(AnAn - 1…A3A2A1)如何证明啊.呼呼 -
咎兰融18697313413 ______[答案] AnAn-1…A3A2A1 =AnAn-1…A6A5A4*1000+ A3A2A1 =AnAn-1…A6A5A4*1000-A3A2A1*1000+1001*A3A2A1 =1000*(AnAn-1…A6A5A4-A3A2A1)+1001*A3A2A1 因为1001A3A2A1=7*143**A3A2A1 所以1001A3A2A1能被七整除 根据已知AnAn-...

通券婷2242数列an是等差数列,若公差d不等于0,a1=1,且a3是a1a9的等比中项,1)求数列an的通项公式.2)若对任意的n属于N+,不等式1/a1a2+1/a2a3+...+1/anan... -
咎兰融18697313413 ______[答案] 1) {an}为等差数列,d≠0 a1=1,a3是a1,a9的等比中项 ∴(1+2d)^2=1+8d ∴4d^2-4d=0 ∵d≠0 ∴d=1 ∴an=n 2) 不等式1/a1a2+1/a2a3+...+1/anan+1≥T 令Sn=1/a1a2+1/a2a3+...+1/anan+1 Sn=1/(1*2)+1/(2*3)+.+1/[n(n+1)] =1-1/2+1/2-1/3+...+1/n-1/(n+1) ...

通券婷2242(2014•内江)如图,已知A1、A2、A3、…、An、An+1是x轴上的点,且OA1=A1A2=A2A3=…=AnAn+1=1,分别过点A1、A2、A3、…、An、An+1作x轴的... -
咎兰融18697313413 ______[选项] A. n+1 2n+1 B. n2 3n−1 C. n2 2n−1 D. n2 2n+1

通券婷2242等差数列{an}中,a2=4,a4+a7=15.(Ⅰ)求数列{an}的通项公式;(Ⅱ)设bn=1anan+1,求b1+b2+b3+…+b10的值. -
咎兰融18697313413 ______[答案] (I)设等差数列{an}的公差为d.由已知得a1+d=4(a1+3d)+(a1+6d)=15解得a1=3d=1,所以an=a1+(n-1)d=n+2;(Ⅱ)∵an=n+2,∴bn=1anan+1=1(n+2)(n+3)=1n+2-1n+3,∴b1+b2+…+b10=13-14+14-15+…+112-113=13-113=1039...

通券婷2242已知数列{an}的通项公式an=2n+1,求{1anan+1}前n项的和. -
咎兰融18697313413 ______[答案] ∵an=2n+1, ∴ 1 anan+1= 1 (2n+1)(2n+3)= 1 2( 1 2n+1- 1 2n+3). ∴{ 1 anan+1}前n项的和= 1 2[( 1 3- 1 5)+( 1 5- 1 7)+…+( 1 2n+1- 1 2n+3)]= 1 2( 1 3- 1 2n+3)= n 6n+9.