An Bn=n 2n-1 则a10 b13

n=1/an=1/(n^2+3n+2)=1/[(n+1)(n+2)]=1/(n+1)-1/(n+2)S10=b1+b2+...+b10=1/2-1/3+1/3-1/4+...+1/11-1/12=1/

因an=2^n,bn=2n-1所以anbn=（2n-1)2^n所以tn=a1b1+a2b2.+anbn=1*2+3*2^2+5*2^3+.+(2n-3)2^(n-1)+（2n-1)2^n两边乘以2得2

n=1/(n2+3n+2)=1/((n+1)(n+2))S10=1/(2*3)+1/(3*4).+1/(11*12)=1/2-1/3+1/3-.+1/11-1/12=1/2-1/12=5/12

an=(n+1)(n+2)anbn=1bn=1/an=1/[(n+1)(n+2)]=[(n+2)-(n+1)]/[(n+1)(n+2)]=(n+2)/[(n+1)(n+2)]-(n+1)/[(n+1)

an=n^2+3n+2=(n+1)(n+2)bn=1/[(n+1)(n+2)]=1/(n+1)-1/(n+2)S10=b1+b2+..+b10=(1/2-1/3)+(1/3-1/4)+..+(1/11

设M为{bn}的上界则|bn|

（1）设数列{an}的公差为d，数列{bn}的公比为q，则由题意知a1b1=1(a1+d)(b1q) =4(a1+2d)(b1q2) =12 ，因为数列{an}各项为正数

∵anbn＝2an2bn＝a1+a2n−1b1+b2n−1=(2n−1)(a1+a2n−1) 2(2n−1)(b1+b2n−1) 2＝s2n−1T2n−1∴anbn=2(2n−1)

1）∵a2=b2∴1+d=1×q∵a4=b4∴1+3d=1×q^3组合成方程组后把d=q-1带入1+3d=q^3q^3-3q+2=0q^3-3q+3-1=0q^3-1-3(q-1)=0(q-1)(q^

an=a1+(n-1)dbn=b1+(n-1)Da1=36.b1=64,a100+b100=100所以d+D=0an的等差为d.则bn的等差为-d数列an+bn是等差为0的等差数列100*200=20

令Tn为{anbn}的前n项和，那么：Tn=a1b1+a2b2+…+anbn=1×20+3×21+5×22+…+（2n-1）•2n-12Tn=1×21+3×22+5×23+…（2n-1）•2n∴Tn=

n是（1/2）n还是1/（2n）

cn=anbn=(3n-1)*2^nSn=2*2^1+5*2^2+……+(3n-1)*2^n2Sn=2*2^2+……+(3n-4)*2^n+(3n-1)*2^(n+1)相减：Sn=(3n-1)*2^(

等我算算啊,几分钟

(n^2+n)x^2-(2n+1)x+1=0由根与系数的关系x1+x2=(2n+1)/(n^2+n)x1x2=1/(n^2+n)|AnBn|=|x1-x2|=√(x1-x2)^2=√[(x1+x2)^

由等差数列的性质和求和公式可得：anbn=2an2bn=a1+a2n−1b1+b2n−1=(2n−1)(a1+a2n−1)2(2n−1)(b1+b2n−1)2=A2n−1B2n−1=7(2n−1)+4

因为Sn=2^n-1所以S(n-1)=2^(n-1)-1所以an=Sn-S(n-1)=2^(n-1)(n>=2)因为S1=a1=2^1-1=1=2^0所以an=2^(n-1)(n>=2)因为bn=n所

∵SnTn=n2n+1，∴a7b7=2a72b7=132(a1+a13)132(b1+b13)=S13T13=132×13+1=1327，故选：C．

答案

∵数列{an}的通项公式是an=2n2n+1=2n+1−12n+1=1-12n+1，（n∈N*），显然当n增大时，an的值增大，故数列{an}是递增数列，故有an＜an+1，故选B．