a1不等于0，2an-a1=S1·Sn

令1/an=bn则（1/an+1）=2+（1/an）可改写成bn+1=2+bna1=3b1=1/3所以bn=（6n-5）/3所以an=3/（6n-5）

如果您满意我的回答,手机提问的朋友在客户端右上角评价点【满意】即可!再问：你的图看不到再答：n=1时，2a1-a1=a1=S1S1=a1²a1²-a1=0a1(a1-1)=0a1=

因为Sm/Sn=m^2/n^2,所以{[2a1+(m-1)d]*m}/{[2a1+(n-1)d]*n}=m^2/n^2,[2a1+(m-1)d]/[2a1+(n-1)d]=m/n,2a1n+(m-1)

因为Sn+1=4an+2Sn=4an-1+2故an+1=4an-4an-1an+1-2an=2（an-an-1）令an+1-2an=bn+1故bn+1=2bn所以bn=b2*2^（n-2）=（a2-2

n=1时,2a1-a1=S1×S1=a1²a1²-a1=0a1(a1-1)=0a1=0(与已知矛盾,舍去)或a1=12a2-a1=S2=a1+a2a2=2a1=2×1=2S1=a1

答案如图

a1+d=a2=b1*q=b2b1q-a1-d=0b1=a1不等于a2,则q不等于1a1=d/(q-1)因an>0,则d>0,否则,总有an小于0的时候.0b3a4=a3+db4=b3*qa4>a3b

猜测,an=1/[n(n-1)]n>1n=2时,a2=1/2假设k=n时,an=1/[n(n-1)]当k=n+1时,Sn=1+1/2*1+…+1/[n(n-1)]=2-1/nan+1=(Sn+an+1

a1=aa2=1/a^2a3=a^4a4=1/a^8……a1*a2…*a10=1/a(1+2^2+2^4+2^6+2^8)=1/a^341

a(n+1)=2an/1+an,1/a(n+1)=1/2an+1/2,1/a(n+1)-1=1/2*(1/an-1),[a(n+1)-1]/a(n+1)=1/2*(an-1)/an所以(an-1)/a

解析：∵a1=4,a7=4+6d,a10=4+9d∴a7^2=a1*a10,即（4+6d)^2=4(4+9d)∵d≠0∴d=-1/3即a1=4,a7=2,a10=1∴q=a2/a1=1/2∴Sn=4*

这是因为q=a(n-1)分之an（n≥2）所以：an=a(n-1)*q所以：a1*a(n-1)*q=a1*an

是A1,A3,A4等比数列吧?∵A1,A3,A4等比数列∴(a3)²=(a1)×(a4)(a1＋2d)²=(a1)(a1＋3d)a²₁+4d²+4a

（1）S1=a1=（2a1/a1）-1=1S2=2a2/a1-1=2a2-1=a1+a2=1+a2所以2a2-1=1+a2a2=2（2）Sn=(2an/a1)-1=2an-1Sn-1=（2an-1/a

a_(n+1)=Ca_n+(2n+1)C^(n+1)a_(n+1)-((n+1)^2)*C^(n+1)=Ca_n-(n^2)C^(n+1)=C[a_n-(n^2)C^n]所以{a_n-(n^2)C^n

S(n+1)=2Sn+a1.(1)Sn=2S(n-1)+a1.(2)(1)-(2)得S(n+1)-Sn=2[Sn-S(n-1)]a(n+1)=2an∴an是q=2的等比数列an=a1X2^(n-1)S

a[n+1]-a[n]=2a[n+1]a[n]1/a[n]-1/a[n+1]=21/a[n+1]=(1/a[n])-21/a[n]为等差数列,公差为-2,首项1/a[1]=1/2所以1/a[n]=1/

T=[2357];a=[1379];fun=@(a,T)a.*ones(1,T);S=cell2mat(arrayfun(fun,a,T),'un',false)再问：我刚刚跑了下你的程序>>T=[2

a(n)=1+(n-1)da(n+1)=1+ndSn=(1+an)n/2=(2+nd-d)n/2(1+Sn)/(n(1-a(n+1)))=-((4+nd-d)/n)/(2n(nd))=-2/(nd)-