F138bn=3×21提示:既然bn=a+1-2an则b1=a2-2a1=5-2×1=3(a2可令条件式n=1求出),bn+1=an+2-2a+1.考虑到bn只与an有关,故条件式中Sn多余,设法消去(这句台词是不是耳朵都听起茧子啦?没错,多写真题就会发现:真题是不断重复的,这才是高考真题的真相)观察:Sn+1=4an+2①,当n≥2时,有S...
A.-4 B.-2 C. 2 D. 4 相关知识点: 试题来源: 解析 答案见上8.A 解析:S .S .S 成等差数列 ∴2Sn=S3+S . ∴(S S )+(Sn-S4)=0, (a_i+a_s+a_i)+(a_i+a_(si)+a_i)+(a_(ij)+a_i) 即(a7+as+an)+(a7 an )=0. ∴2(a_1+a_1+a_1+a_1+a_1+a_1+a_1+a...
Sn1per is anautomated scannerthat can automate the process of collecting data forexploration and penetration testing. In their work sn1per involves such well-known tools like: amap, arachni, amap, cisco-torch, dnsenum, enum4linux, golismero, hydra, metasploit-framework, nbtscan, nmap smtp-user...
The difference between SN1 and SN2 reaction with example: In an SN1 reaction, 2-bromo-2-methylpropane reacts via a carbocation intermediate, while in an SN2 reaction, methyl chloride reacts with hydroxide in a single step. 6. Can you explain the difference between SN1 and SN2? The differ...
3.已知首项为3的数列\(a_n\) 的前n项和为Sn,且 Sn1+an=S_n+5⋅4^n .(1)求证:数列 \(a_n-4^n\) 为等比数列.(2)求数列{an}
1)因为 Sn.1 =Sn+an+1(nEN'),所以 Sn.1-Sn=an+1.即an.1=an+1.所以an. -a.= 所以|an}是以 a_1=1 为首项,} 为公差的等差数列,所以 a_(tr)=u , 设正项等比数列bn}的公比为q(q0),由 b_2b_4=b_1^2=1 ,解得 b_3=1 , 又、是1 2 6,与5b、的等差中项,所以 12b_1...
An example of an aromatic nucleophilic substitution reaction is the Bamberger rearrangement. Electrophilic nitrogen and oxygen-containing side group, and benzene work together to add a hydroxyl (OH) group to the benzene chain. That chemical formula is: (C6H6)NHOH + H2SO4/H2O -> HO(C6H6)NH2...
ALLEN BRADLEY 50HP VARIABLE SPEED AC DRIVE 1336S-B050-AN-EN4-L6 SER. D ALLEN BRADLEY 505-BAD 505BAD ALLEN BRADLEY 503352 503352 Allen Bradley 5 HP Drive 1336SBRF50ANEN5 Alt Part # 1336SBRF50ANEN5HA2L6 E ALLEN BRADLEY 445L-P4S0480YD 445LP4S0480YD ...
S_(n-1)=(a_n)/(λa_n-1) " 两式相减得,an= a_n=(a_(n+1))/(λa_(n+1)-1)-(a_n)/(λa_n-1)⋯a_n+(a_n)/(λa_n-1)=\frac(a_n dn+1 n I 2 + Aan-1 Aan+-1 (λn_n)/(λ_n-1)=(a_n+1)/(λa_n+1-1),(λa_n-1)/(λa_n^2)=(λn...
解析 解析 依题意,当 n≥2 ,Sn=a1(Sn+1)①, i_n=a_1(S_(u-1)+1) ②①-②41/1 an =a_1a_n=2a .因为S =2(S_1+1) ,即 a_1+a_2=2(a_1+1) ,所以 a_2=4 ,即 a_2=2a_1 ,所以 a_n=2^n(n≥2) . ∴ 所以a =2^⋅(n∈N^*) . ...