记sn是各项均为正数的等差数列

√S2=√S1+d=√a1+d,S2=（√a1＋d）²=a1+d²+2d√a1a2=S2-S1=a1+d²+2d√a1-a1=d²+2d√a1a3=2a2-a1

这是今年江苏卷上的题目…………(1)设根号Sn=d*n+HSn=d^2*n^2+2*d*H*n+H^2a1=S1=d^2+2*d*H+H^2a2=S2-S1=3*d^2+2*d*Ha3=S3-S2=5

1、√S1=√a1√S2=√(a1+a2)=√a1+2(1)√S3=√(a1+a2+a3)=√(3a2)=√a1+4(2)由(1)得a1+a2=a1+4√a1+4√a1=(a2-4)/4代入(2)√(

设根号sn=x+2(n-1)sn=(x+2(n-1))²an=sn-s(n-1)an=(x+2(n-1))²-(x+2(n-2))²an=(2x+2(n-1)+2(n-2

√Sn=√S1+(n-1)d√S2=√S1+d√S3=√S1+2d第2个式子两边平方a1+a2=a1+2(√a1)d+d^2第3个式子两边平方a1+a2+a3=a1+4(√a1)d+4d^2两个式子相

s1=a1;s2=a1+a2;s3=a1+a2+a3=3a2根号s3=根号s1+2d=根号s2+d化简得a2=3a1代入等差数列可求得d=根号a1sn=(nd)²an=sn-sn-1=(2n

结果是an=4(2n+1);首先由s1,s2,s3的关系可列出两个方程,关于a1,a2,a3.和已知的2a2=a1+a3联立,求出a1=4.接下来,利用根号sn是等差数列,推导出s（n）和a1的关系,

2S2=b2(a1+a2)=b1*q*(2a1+d)=32,b3S3=b3(a1+a2+a3)=b1*q²*(3a1+3d)=120,得d=2(都是正数）,q=2.∴an=a1+d(n-1)

1.n=1时,2a1=2S1=a1²+1-4a1²-2a1-3=0(a1+1)(a1-3)=0a1=-1(数列各项均为正,舍去)或a1=3n≥2时,2an=2Sn-2S(n-1)=

sn=an(an+1)/2s(n-1)=a(n-1)(a(n-1)+1)/2两式相减an=an(an+1)/2-a(n-1)(a(n-1)+1)/2an^2-an-a^2(n-1)-a(n-1)=0(

n=1时,2a1=2S1=a1^2+1-4a1^2-2a1-3=0(a1+1)(a1-3)=0a1=-1(数列各项均为正,舍去)或a1=3n≥2时,2an=2Sn-2S(n-1)=an^2+n-4-a

此题是数列和均值不等式的结合S10=(a1+a10)/2*10=15a1+a10=3∴a3+a8=a1+a10=33[(1/a3)+(4/a8)]=(a3+a8)(1/a3+4/a8)=1+4+a8/

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

sn=(1/8)(an+2)²S(n-1)=(1/8)[a(n-1)+2]²an=Sn-S(n-1)=(1/8){(an+2)²-[a(n-1)+2]²}=(1

∵{log2an}是公差为-1的等差数列∴log2an=log2a1-n+1∴an=2log2a1−n+1=a1•2−n+1∴S6=a1（1+12+…+132）=a1•1−1261−12=38，∴a1

根号Sn的通项公式是nSn=n^2an=Sn-Sn-1=n^2-(n-1)^2=2n-1

Sn、an、1成等差,则2an=Sn＋1(n=1时,得a1=1),当n≥2时,有2a(n－1)=S(n－1)＋1,则2an－2a(n－1)=an,即an/[a(n－1)]=2=常数,所以{an}是等比

由题意知2an=Sn+1/2,an＞0,当n=1时,2a1=a1+1/2,解得a1=1/2,当n≥2时,Sn=2an-1/2,S(n-1)=2a(n-1)-1/2,两式相减得an=Sn-S(n-1)=

（1）由Sn，an，12成等差数列，可得2an＝Sn+12，∴a1＝12，a2=1（2）由2an＝Sn+12可得，2Sn=4an-1（n≥1），∴2Sn-1=4an-1-1（n≥2）∴两式相减得2an

由题意2an=Sn+1/2Sn=2an-1/2n=1时,S1=a1a1=2a1-1/2a1=1/2S(n+1)-Sn=a(n+1)2a(n+1)-1/2-[2an-1/2]=a(n+1)a(n+1)=