设sn是正数数列an的前n项和,且sn=1 4an^2 1 2an-3 4

∵an与2的等差中项等于Sn与2的等比中项，∴12(an+2)＝2Sn，即Sn＝18(an+2)2．  …（2分）当n=1时，S1＝18(a1+2)2⇒a1＝2； …（3

(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

an+Sn=4a(n-1)+S(n-1)=4相减：an/a(n-1)=1/2等比数列n=1时a1+a1=4a1=2an=2^(2-n)bn=1/n²数学归纳法n=2时T2=5/4

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

设公比为q,则q>0a3=a2+4a1q^2=a1q+4a1=2代入,整理,得q^2-q-2=0(q+1)(q-2)=0q=-1(舍去)或q=2Sn=a1(q^n-1)/(q-1)=2×(2^n-1)

1）由题意得,a1=1,当n>1时,sn=an^2/2+an/2sn-1=a(n-1)^2/2+a(n-1)/2,∴sn-sn-1=an^2/2-a(n-1)^2/2+an/2-a(n-1)/2即（a

(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

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

【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=

再答：再问：求解答～已知数列｛an｝满足a1+a2/2+…+an/n=2∧n-1(n∈N),求数列an的通项公式，再答：本人已死再问：已知函数f(x)=x³+ax²-a²

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

an与1的等差中项为：(an+1)/2因为{an}是正数组成的数列,所以Sn与1的等比中项为根号Sn那么根号Sn=(an+1)/2所以Sn=(an+1)^2/4当n1=,a1=(a1+1)^2/4即a

由已知an与1的等差中项等于Sn与1的等比中项得(an+1)/2=√SnSn=(an+1)²/4n=1时,S1=a1=(a1+1)²/4,整理,得(a1-1)²=0a1=

因为an与2的等差中项等于Sn与2的等比中项所以(an+2)/2=√(2Sn)即Sn=(an+2)^2/8.(1)当n=1时a1=S1=(a1+2)^2/8解得a1=2当n≥2时S(n-1)=(a(n

[a(n)+2]^2=8s(n),[a(1)+2]^2=8s(1)=8a(1),[a(1)-2]^2=0,a(1)=2.[a(2)+2]^2=8s(2)=8[a(1)+a(2)],[a(2)-2]^2

1.8A1=8S1=(A1+2)^2(A1)^2-4A1+4=0A1=28(A1+A2)=8S2=(A2+2)^2(A2)^2-4A1-12=0A2=6A2=-2（舍去）8(A1+A2+A3)=(A3

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

楼上的你的已经错了好不好啊bn=4/(an*an+1)=1/(4n-2)-1/(4n+2)错了!应该是bn=4/(an*an+1)=4/an-4/an+1Tn=4/a1-4/an+1不要误人子弟好不好

由题意得(an+1)/2=√(Sn×1)Sn=[(an+1)/2]²n=1时,S1=a1=[(a1+1)/2]²,整理,得(a1-1)²=0a1=1n≥2时,Sn=[(a

a1=2,a2=6,a3=10(an+2)/2=√2sn(an+2)^2=8sn(a(n-1)+2)^2=8s(n-1)相减：(an+2)^2-(a(n-1)+2)^2=8sn-8s(n-1)an^2