正数数列an的前n项和sn,满足2根号下sn=an 1

（1）∵2Sn=an2+n-4（n∈N*）．∴2Sn+1=an+12+n+1-4．两式相减得2Sn+1-2Sn=an+12+n+1-4-（an2+n-4），即2an+1=an+12-an2+1，则an

好像问题没写全吧

∵Sn-Sn-1=√Sn+√Sn-1∴(√Sn)²-(√Sn-1)²=√Sn+√Sn-1(√Sn-√Sn-1)(√Sn+√Sn-1)=√Sn+√Sn-1∴√Sn-√Sn-1=1（n

(1)S[1]=a[1]=1/2(a[1]+1/a[1]),于是：a[1]=1=√1-√0S[2]=a[2]+1=1/2(a[2]+1/a[2]),于是：a[2]=√2-1,S[2]=√2S[3]=a

证明：∵Sn=an(an+1)2∴S1=a1(1+a1)2∴a1=1…（1分）由2Sn=a2n+an2Sn-1=a2n-1+an-1⇒2an=2(Sn-Sn-1)=a2n-a2n-1+an-an-1…

1楼貌似错了!（a1^2-3a1=6a1与An^2+3An=6Sn矛盾）An^2+3An=6SnA(n+1)^2+3A(n+1)=6S(n+1)后减前得A(n+1)^2+3A(n+1)-An^2-3A

2√Sn=an+1则有,4Sn=(an+1)²4a(n+1)=4[S(n+1)-Sn]=[a(n+1)+1]²-(an+1)²=[a(n+1)]²+2a(n+1

n=1,a1=2,n>1,an-1+Sn-1=4(1)an-Sn=4(2)(2)-(1)an/an-1=1/2,an=2x(1/2)^(n-1)=0.5^(n-2)n为正整数(2)补充问题你在检查一下

(1)(an+2)/2=根号下2Sn所以8Sn=(an+2)^2n=1,S1=a1.8a1=(a1+2)^2,得a1=2n=2,8S2=(a2+2)^2,8(a1+a2)=(a2+2)^2,得a2=6

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)=

1）n=1,解得a1=1n>1时S(n-1)=1/4(a(n-1)+1)^2Sn=1/4(an+1)^2相减并整理得到an^2-2an-a(n-1)^2-2a(n-1)=0(an-a(n-1)-2)(

4a(1)=[a(1)+1]^2a(1)=14a(n+1)=[a(n+1)+1]^2-[a(n)+1]^2[a(n)+1]^2=[a(n+1)-1]^2若a(n+1)>1a(n+1)=a(n)+2a(

【解法一】Sn=1/2(an+1/an)S(n-1)=Sn-an=1/2(1/an-an)Sn+S(n-1)=1/anSn-S(n-1)=an上面两式相乘得：Sn^2-S(n-1)^2=1S1=a1=

当n=1时,S1=a1=1/2(a1^2+a1),解得a1=1当n＞1时,an=Sn-S(n-1)=1/2(an^2+an)-1/2[a(n-1)^2+a(n-1)],整理得[an+a(n-1)][a

因为6Sn=(an+1)(an+2)（1）所以6Sn-1=(an-1+1)(an-1+2)（2）（1）-（2）则an-an-1=3所以an是等差数列因为6Sn=(an+1)(an+2)可知S1＝a1=

（1）a1＝(a1+1)24，解得a1=1，当n≥2时，由an=Sn-Sn-1=(an+1)2−(an−1+1)24，得（an-an-1-2）（an+an-1）=0，又an＞0，所以an-an-1=2

（1）当n=1时，a1＝s1＝14a21+12a1−34，解出a1=3，又4Sn=an2+2an-3①当n≥2时4sn-1=an-12+2an-1-3②①-②4an=an2-an-12+2（an-an

2根号Sn=an+14Sn=an的平方+2an+14Sn_1=an_1的平方+2an_1+1〔n≥2〕又Sn-Sn_1=an所以4an=an的平方+2an-an_1的平方-2an_1划简为〔an+an

1.4a1=4S1=(a1+1)²整理,得(a1-1)²=0a1=14S2=4a1+4a2=4+4a2=(a2+1)²整理,得(a2-1)²=4a2=-1(舍去

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