等差数列an和bn中,a11 b11=4 3

an,（bn)^2,a(n+1)成等差数列2(bn)^2=an+a(n+1)--①由（bn）^2,a(n+1),(b(n+1))^2成等比数列(a(n+1))^2=[bnb(n+1)]^2∴a(n+1

a100+b100=a1+b1=100a1+b1+99d1+99d2=a1+b1可知2者的公差互为相反数所以an+bn=a1+(n-1)d1+(n-1)d2+b1=a1+b1=100所以{an+bn}

An=[2n/(3n+1)]BnAn-1=[2n/(3n+1)]Bn-1lim(n→∞)an/bn=lim(n→∞)[An-An-1]/[Bn-Bn-1]=lim(n→∞)[2n/(3n+1)][Bn

（1）由条件得2bn=an+an+1，an+12=bnbn+1由此可得a2=6，b2=9，a3=12，b3=16，a4=20，b4=25…（6分）（2）猜测an=n（n+1），bn=（n+1）2用数学

根据bn=1/(an*a(n+1)),我们知道,bn=[1/an-1/a(n+1)]/d.因此,Sn=[1/a1-1/a2+1/a2-1/a3+...+1/a(n-1)-1/an]/d=[1/a1-1

a2a3a4=15则a3=5a4=(59)÷2=7则公差d=2则a2=3,a1=1,an=2n-1bn=根号3×(1an)bn=2n×根号3b1=2根号3,b2=4根号3,b3=6根号3,则公差d=2

d=(a5-a2)/(5-2)=9/3=3a1=a2-d=6-3=3所以an=3+3(n-1)=3nbn=a(2n)=6n所以前5项和=（6+30）*5/2=18*5=90再问：怎么知道｛bn｝是等差

由已知,得,公差d=(a5-a2)/3=3所以a1=a2-d=6-3=3所以an=3nbn=a2n=6n所以{bn}是以6为首项,6为公差的等差数列所以数列｛bn｝的前5项和=（6+30）*5/2=9

等差数列,所以an=a1+(n-1)dy由a2=9,a5=21,可以根据上面的式子算出a1=5,d=4所以an=4n+1所以bn=2^4n+1bn+1/bn=2^4(n+1)+1/2^(4n+1)=2

在等差数列｛an｝中,a1+a3=6,a11=21,可解得a1=1,d=2.∴an=2n-1∴bn=1/n（an+3）=1/[n(2n+2)]=[(1/n)-1/(n+1)]/2∴Sn=b1+b2+.

a(n+1)=√[bn*b(n+1)]2bn=an+an+12bn=√[bn*b(n-1)]+√[bn*b(n+1)]2√bn=√b(n-1)+√b(n+1)所以数列{√bn}为等差数列√b1=√2(

a(n+1)=2an+2^na(n+1)/2^n=2an/2^n+1a(n+1)/2^n=an/2^(n-1)+1a(n+1)/2^n-an/2^(n-1)=1,为定值.a1/2^(1-1)=1/1=

a2+a6=2a4=14a4=7公比d=a5-a4=9-7=2an=a4+d(n-4)=7+2(n-4)=2n-1bn=an+2^n=2n-1+2^nSn=(2+2n)*n/2-n+2(1-2^n)/

1=a2=5b2=4a=9所以q=b2/b1=9/5Sn=b1(1-q^n)/(1-q)=(25/4)[(9/5)^n-1]

(1)a1=2,b1=42*4=2+a2,则a2=66^2=4*b2,则b2=92*9=6+a3,则a3=1212^2=9*b3,则b3=16由a1=2=1*2,a2=6=2*3,a3=12=3*4猜

(2)由已知得an=n(n+1),bn=(n+1)^2,所以an+bn=2n^2+3n+1>2n^2+2n=2n(n+1),所以1/an+bn

由a2＝a1+d＝6a5＝a1+4d＝15⇒a1＝3d＝3∴an=3+3（n-1）=3n       bn=a3n=9n∴S9＝9

设{an}的公差为d，首项为a1，由题意得a1+d＝6a1+4d＝15，解得a1＝3d＝3；∴an=3n，∴bn=a2n=6n，且b1=6，公差为6，∴S5=5×6+5×42×6=90．故选C．

a1=b1a2n+1=b2n+1a2n+1-a1=b2n+1-b1=2nd=b1(q^2n-1)=b1(q^n+1)(q^n-1)nd=b1(q^n+1)(q^n-1)/2an+1-bn+1=a1+n

a(n+1)=√[bn*b(n+1)]2bn=an+an+12bn=√[bn*b(n-1)]+√[bn*b(n+1)]2√bn=√b(n-1)+√b(n+1)所以数列{√bn}为等差数列2.√b1=√