正数列an的前n项和为sn且2根号Sn=an 1

当n=1时,2S1=a1+1/a1,得a1=1当n=2时,2S2=2(1+a2)=a2+1/a2,得a2=√2-1当n=3时,2S3=2(√2+a3)=a3+1/a3,得a3=√3-√2猜想an=√n

（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

(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

因为2√S（1）=2√a（1）=a（1）+1所以a（1）=1因为2√S（n）=a（n）+12√S（n+1）=a（n+1）+1以上2式分别平方,再相减,得：4·a（n+1）=[a（n+1）]^2+2·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…

Sn是an^2和an的等差中项所以Sn=（an²+an）／2①同理得Sn-1=（an-1²+an-1）／2②①-②得2an=an²-an-1²+an-an-1化

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时,S1=a1=1/2(a1+1/a1),a1=1.n=2时,S2=a1+a2=1+a2=1/2(a2+1/a2),a2=√2-1.n=3时,S3=a1+a2+a3=√2+a3=1/2(a3+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)=

由a1^3+a2^3+a3^3+.+an^3=Sn^2得a1^3+a2^3+a3^3+.+an^3+an+1^3=Sn+1^2两式相减得：an+1^3=Sn+1^2-Sn^2=(Sn+1+Sn)(Sn

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

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

【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=

（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

an^2/a(n+1)=an^2/(an×q）=an/qTn=a1/q+……+an/q=Sn/q当Sn

2*Sn^(1/2)=An+1(1)2*S1^(1/2)=A1+1,S1=A1A1=1(2)Sn=(An+1)^2/4S(n-1)=[A(n-1)+1]^2/4An=Sn-S(n-1)=(1/4)*(

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)=