已知an是等差数列且an>0,a2a4 2a3a5 a4a6=25,a3 a5

a2005*a20060,a20050,则a2007+a2006>0因为a2005+a2006=a1+a40100所以使前n项之和sn

an=3n,bn=2^(n-1)分式上下同时乘以2,把2bn化成b(n+1),另s=b(n+1),则cn=s/[（s+1）（s+2）]=s/（s+1）-s/（s+2),另dn=bn/（bn+1）,则c

a1,a2,a3,a4,a5……其中a2=-1,a5=-5可以得出a3=-2.3333……,a4=3.666……也就是以4/3递减a1=1/3,则aN=a1-(N-1)*4/3=-N4/3+5/3Sn

bn=sn-s（n-1）=1-1/3^n-（1-1/3^n-1）=-1/3^n+3/3^n=2/3^n

标题对,还是补充对啊?因为m+n=p+q时,am+an=ap+aq观察下标得4(a1+an)=88sn=n(a1+an)/2=286n=26

不知道你的2^n+1是不是2^(n+1)（1）对an+1-2an=2^n+1两边同时除以2^(n+1)得a(n+1)/2^(n+1)-an/2^n=1因为a1/2=1,所以数列｛an/2^n｝是以1为

当{an}是常数列时,满足题设不满足结论.

sn=an^2+bns(n-1)=a(n-1)^2+b(n-1)两式作差,由：sn-s(n-1)=an可证．

a1*p=a2a1*p^3=a4,a1*p-a1=a1*p^3-a1*Pp-1=p^(p^2-1);(p-1)(p*(p+1)-1)=0,p=1,或p^2+p-1=0,p=(-1+√5)/2,p=(-

a1,a2,a4成等差数列2a2=a1+a4即2a1*q=a1+a1q^3a1不为0所以：2q=1+q^3q^3-2q+1=0q^3-q^2+q^2-2q+1=0q^2*(q-1)+(q-1)^2=0

a1,a2,a4成等差数列所以2a2=a1+a4{an}是等比数列a2=a1qa4=a1q^3所以2×a1q=a1+a1q^3即：q^3-2q+1=0(q-1)(q^2+q-1)=0q=1或q=(-1

a1,a2,a4成等差数列所以2a2=a1+a4{an}是等比数列a2=a1qa4=a1q^3所以2×a1q=a1+a1q^3即：q^3-2q+1=0(q-1)(q^2+q-1)=0q=1或q=(-1

(a3)^2=a13*a1(a1+2d)^2=(a1+12d)*a1d-2a1=0d=2a1s1=a1s3=3a1+3d=9a1s9=9a1+36d=81a1(s3)^2=s1*s9,所以s1s3s9

a5=(a2+a8)/2=(b2+b8)/2;b5=根号（b2*b8）；由基本不等式根号ab==b5q≠1,bi>0a5>b5B

设an公差为d那么通过等差数列定义,只要bn-b(n-1)是常数bn-b(n-1)=an+a（n+1）-[a(n-1)+an]=a（n+1）-a(n-1)=2d所以bn是等差数列.

Sn=n(an+1)/2S(n+1)=(n+1)[a(n+1)+1]/2用下式减上式a(n+1)=[(n+1)a(n+1)-nan+1]/2即2a(n+1)=[(n+1)a(n+1)-nan+1]即(

设an=a1+(n-1)d,bn=an+a(n-1)=a1+(n-1)d+a1+nd=2a1+(2n-1)dbn为首项为2a1-d,公差为2d的等差数列

3d=a5-a2=-5-1=-6d=-2所以a1=a2-d=3a3=a2+d=-1

B(n+1)-Bn=A(n+1)+A(n+2)-An-A(n+1)=A(n+2)-An因为An是等差数列,所以A(n+2)-An=2d是一个与n无关的常数,所以Bn是等差数列

设a2=a,a3=aq,a4=aq^2,a5=aq^3,a6=aq^4a2*a4+2a3*a5+a4*a6=a*aq^2+2aq*aq^3+aq^2*aq^4=a^2(q^2+2q^4+q^6)=a^