已知an是正数组成的数列,其前n项和为Sn

(1)已知a3=4S3=a1+a2+a3---->a1+a2=7-4=3a2*a2=a1*a3------>4a1=a2*a2由1.2可求得a2=2或者a=-6题目已知数列{an}是各项都是正数的等比

∵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

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

因为点（an，an+1）（n∈N*）在函数y=x2+1的图象上，所以an+1＝(an)2+1＝an+1，即an+1-an=1，所以数列{an}是以1为首项，以1为公差的等差数列，则an=a1+（n-1

一、a2a4=1a1qa1q^3=1a1^2q^4=1{an}是由正整数组成的等比数列a1>0q>0a1q^2=1S3=[a1(1-q)^3]/(1-q)=7a1(1+q^2+q)=71+q^2+q=

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

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

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=

/>由已知条件列式：(an+2)/2=√(2Sn)整理,得(an+2)²=8Sn令n=1(a1+2)²=8a1整理,得(a1-2)²=0a1=2令n=2(a2+2)

∵{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

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

1当n=1时易得a1=t(an+t)^2/4=Sn*t展开4t*Sn=(an+t)^24t*S(n-1)=(a(n-1)+t)^2相减,配方an-t=an-1+tan=an-1+2tan=(2n-1)

[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

由a2a4=4,得a3=√4=2,设公比的倒数1/q=t,∵S3=7/2,∴2（1+t+t＾2）=7/2,解得t=1/2（数列各项为正,舍去负的解）q=2∴a1=1/2,a2=1等等,不难得到s5=1