各项均为正数的等比数列an满足a3-a2=2 S3=7 求数列an的通项公式

∵等比数列{an}的各项均为不等于1的正数，数列{bn}满足bn=lnan，b3=18，b6=12，∴a3=a1q2=eb3=e18，a6＝a1q5=eb6=e12，∴a6a3=q3=e12e18=e

（1）∵对任意n∈N*，有Sn＝16(an+1)(an+2)①当n≥2时，有Sn−1＝16(an−1+1)(an−1+2)②当①-②并整理得（an+an-1）（an-an-1-3）=0，而{an}的各

an为等比数列由于bn=log2an,则bn为等差数列,设bn公差为d则b1+b2+b3=3推出3b1+3d=3进而d=1-b1再由题：b1b2b3=-3推出b1^3+3*d*b1^2+2*d^2*b

1、a5-a2=3d=6d=2a1=a2-d=0an=0+2(n-1)所以an=2n-22、b3=a3=2*3-2=4T3=b1+b2+b3=7b1+b2=3b1+b1*q=3b1(1+q)=3b3=

正数项等比数列an/an-1=q,q>0根号an/根号an-1=根号q,所以{根号an}仍是等比数列.

是原数列是a1a1qa1q^2a1q^3a1q^4.根号an根号a1(根号a1)*(根号q)(根号a1)*q(根号a1)*(根号q)*q.任意相邻两项比值为是根号q因为原来q是等比数列公比,根号q不会

是{an}是各项均为正数的等比数列q大于0{根号an}是以根号a1为首项根号q为公比的等比数列

是

.{An}为正数等比数列.那么等比数列的通项公式是：An=A1×q^（n－1）将两边同时开方等式仍然相等.An^1/2=（A1^1/2）×[q^（n－1）]^1/2即

在等比数列中有a5a6=a4a7=a3a8=a2a9=a1a10所以有log3a1+log3a2+...+log3a10=log3(a5a6*a4a7*a3a8*a2a9*a1a10)=5log3a5

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(

∵a2*a4=4∴a3=2.q=1/2.an=2^(4-n)2^(9-3n)>1/9.9-3n>=-3n

a1(q+q^3)=4a1(1+q+q^2)=14两式相除：(q+q^3)/(1+q+q^2)=2/7求得qan+an+1+an+2=（a1+a2+a3）*q^(n-1)>1/9关键是求q说实在的,我

可用递推法：2Sn=An+An*An递推2Sn-1=An-1+An-1*An-1两市相减,得：An+An-1=An*An-An-1*An-1因为An为正数,所以An-An-1=1之后求An,然后用求和

将a(k+1)=(k+1)(k+2)/2b(k+1)=(k+2)^2/2代入已知的等比数列的等差数列看是否成立就行了再问：这样写也就是先假设an=n(n+1)/2成立得到bn吧，那用不用再假设bn成立

刚才做好的突然没了,重来Sn=1/6(An+1)(An+2),a1+a2+a3+...+an=1/6(an+1)(an+2),6(a1+a2+a3+...+an)=(an+1)(an+2),6(a1+

6Sn=An^2+3An+26S(n-1)=[A(n-1)]^2+3A(n-1)+26Sn-6S(n-1)=6An=An^2+3An+2-{[A(n-1)]^2+3A(n-1)+2}An-A(n-1)

1.A(n+1)^2*An+A(n+1)*An^2+A(n+1)^2-An^2=0两边同除以A(n+1)²An²1/An+1/A(n+1)+1/An²-1/A(n+1)&

a1,qb1=log2a1b2=log2a2=loga1+log2qb3=log2a3=log2a1q^2=log2a1+2log2q相加得log2a1q=log2a2=1a2=a1q=2log2a1

（Ⅰ）设数列an的公比为q，则a2＝a1q＝2a4＝a1q3＝12…（2分）解得q＝12，a1＝4（负值舍去）．…（4分）所以an＝a1qn−1＝4•(12)n−1＝2−n+3．…（6分）（Ⅱ）因为a