Sn是其前n项和,若 S2017 2017

由Sn^2=an(Sn-1/2),两边同时除以Sn,拆开括号,得到Sn=an-an/2Sn,移项,an-Sn=an/2Sn,两边同时除以an,乘以2,得到2(an-Sn)/an=1/Sn,那么1/(S

设{an}公差为d,{bn}公比为q,q>0a3+b3=a1+2d+b1q^2=1+2d+3q^2=17d=(16-3q^2)/2T3-S3=b1(1+q+q^2)-(a1+a1+d+a1+2d)=3

Sn=4(4^n-1)/(4-1)-2(2^n-1)/(2-1)=[4^(n+1)-4)/3-[2^(n+1)-2]=[4^(n+1)-4-3*2^(n+1)+6]/3=[2^(n+1)*2^(n+1

解题思路：根据Sn与an的关系可以解解题过程：varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include

解,a5/b5=65/12解法1,a5/b5=S9/T9=65/12解法2,Sn/Tn=7n+2/n=3s(2n-1)/T(2n-1)=7(2n-1)+2/2n-1+3=14n-5/2n+2an/bn

（1）当n=1时,S2=2µ*S1+1=2µ*a1+1,S2=2µ+1当n=2时,S3=2µ*S2+1,则S2+a3=2µ*S2+12&mi

由题意a3=16,故S5=5×a3=80,由数列的性质S10-S5=80+25d,S15-S10=80+50d,S20-S15=80+75d,故S20=20=320+150d,解之得d=-2又S10=

(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

/>an=sn-s(n-1)=an+n^2-n-[a(n-1)+(n-1)^2-(n-1)]=an-a(n-1)+2n-2a(n-1)=2(n-1)an=2n所以,数列an为公差为2的等差数列.

a3=a1+2d=6S3=a1+a2+a3=3a1+3d=12解得a1=2,d=2,故an=2n所以Sn=n(n+1)所以1/S1+1/S2+……+1/Sn=1/(1*2)+1/(2*3)+1/(3*

S17=a1+a2+……+a17=17a9T17=b1+b2+……+b17=17b9(先利用等差数列的特性（n项相加等于它的中位数）,再用等量置换的方法)Sn/Tn=2n+3/3n-1S17/T17=

1.证：Sn=(3an-n)/2Sn-1=[3a(n-1)-(n-1)]/2an=Sn-Sn-1=[3an-3a(n-1)-1]/2an=3a(n-1)+1an+1/2=3a(n-1)+3/2=3[a

若S10>0,则S10=（a1+a10)*10/2>0则2a1+9d>0.则d>-2a1/9同理S11

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

(1)q=1a(n+1)/Sn=1/nlim(an+1/Sn)=lim(1/n)=0=1-1∴q=1满足(2)q≠1Sn=a1(1-q^n)/(1-q)a(n+1)=a1*q^na(n+1)/Sn=[

因为Sn=3n^2+5nS（n-1）=3（n-1）^2+5(n-1)两式相减所以an=6n-3+5=6n+2所以an=8+6（n-1）,所以an是以8为第一项,公差为6的等差数列.

由题意有：S3=a1（1+q+q^2）=3*a1*q^2即：1+q+q^2=3q^2所以：2q^2-q-1=0即（2q+1）(q-1)=0解得：q=-1/2或q=1

（1）令n=1,得a1=-1.Sn=2an+n,S(n+1)=2a(n+1)+n+1.两式相减,得a(n+1)=2a(n+1)-2an+1.整理得a(n+1)-1=2(an-1),a1-1=-2.综上

（1）3an=2Sn+n...①3an+1=2Sn+1+n+1...②②-①得：3an+1-3an=2an+1+1即an+1=3an+1==＞an+1+1/2=3(an+1/2)an+1+1/2/an

∵{an},{bn}是两个等差数列∴a1+a15=2a8b1+b15=2b8∴a8/b8=(15(a1+a15)/2)/(15(b1+b15)/2)=S15/T15∵Sn/Tn=(7n+2)/(n+3