6Sn=an2 3an

分析：由于对于数列的n值有不同范围取值,对应不同的求和公式,可知数列为分段数列,需要对不同范围的n值进行讨论,方可求得数列的通项公式；当n=1时,a1=S1=3+1=4;当2≤n≤5时,an=Sn-S

S(n-6)-S6=a7+a8+.+a(n-6)=[a7+a(n-6)]*(n-12)/2=[a1+a(n)]*(n-12)/2=108(1)Sn=a1+a2+.+a(n)=[a1+a(n)]*n/2

S6=(a1+a6)*6/2=362a1+5d=12Sn-S(n-6)=180即[a(n-5)+an]*6/2=180最后6项的和是6an-15d=1802an-5d=60相加2(a1+an)=72S

∵Sn=324，Sn-6=144，∴Sn-Sn-6=an-5+an-4+…+an=180又∵S6=a1+a2+…+a6=36，a1+an=a2+an-1=a6+an-5，∴6（a1+an）=36+18

(5n-8)Sn+1-(5n+2)Sn=A*n+B合并同类项,得10Sn=1-A*n-B因为S1=a1,S2=a1+a2=7,所以10S1=1-A-B,即10=1-A-B.110S2=1-2A-B,即

Sn-S(n-m)=A(n-m+1)+A(n-m+2)+……+A(n-m+m)=b共m项A(n-m+1)=A1+(n-m)dA(n-m+2)=A2+(n-m)d……A(n-m+m)=An=Am+(n-

由Sn=Sn-1/2Sn-1+1,两边同时取倒数可得1/Sn=(2Sn-1+1)/Sn-11/Sn=2+1/Sn-1即1/Sn-1/Sn-1=2故{1/Sn}是首项为1/2,公差为2的等差数列1/Sn

q=(a2+a3+a4)/(a1+a2+a3)=-1/2a1+a2+a3=a1+a1*q+a1*q*q=6则a1=8sn=a1(1-q的n次方）/1-q因为0

因为Sn=324,s(n-6)=144所以最后六项和=324-144=180=a(n-5)+a(n-4)+,+an又S6=36=a1+a2+,+a6两侧同时相加,有6(a1+an)=216a1+an=

Sn-Sn-6=an+an-1+an-2+...+an-5=a6+(n-6)d+a5+(n-6)+...+a1+(n-6)d=S6+6(n-6)d所以d=24/(n-6)S6=3(a1+a6)=3(2

等比数列(1)a1=3,q=2,n=6Sn=a1(1-q^n)/(1-q)S6=3*(1-2^6)/(1-2)=3*(2^6-1)=3*63=189(2)an=a1*q^(n-1)1/2=8*(1/2

等差数列前n项和Sn=na1+n*（n-1）*d/2n=6时S6=6a1+6*5*d/2S6=6a1+15d36=6a1+15da1=6-(5/2)dSn=na1+n*(n-1)*d/2=324将a1

条件不足,无解,但注意高级魔法师的前n相和的求法只适用于等比数列公比小于1的情况,和此题无关

a1=1s1=1s(n+1)=sn+a(n+1)=sn+an+6,a(n+1)=an+6是个等差数列an=1+6*(n-1)=6n-5

由a1=S1=1/6(a1+1)(a1+2),解得a1=1或a1=2,由假设a1=S1>1,因此a1=2,又由a(n+1)=S(n+1)-Sn=1/6(a(n+1)+1)(a(n+1)+2)-1/6(

An=6Sn/(An+3)6Sn=(An)^2+3Ann>=26S(n-1)=(A(n-1))^2+3A(n-1)6An=(An)^2+3An-(A(n-1))^2-3A(n-1)(An)^2-(A(

Sn是等差数列An的前n项和,已知S6=36,Sn=360,Sn-6=216,求项数n【解】由S6=(a1+a6)*6/2=36,得a1+a6=12,记为1式Sn-S(n-6)为后6项的和,由Sn-S

 再问： 再问：那个划横线的答案是不是错了再答：我觉得是

a1=1,a2=6,Sn=3S(n-1)-2S(n-2)+n^2Sn-S(n-1)=2S(n-1)-2S(n-2)+n^2an=2a(n-1)+n^2a1=1,a2=2*1+2^2=6a3=2*6+3

S6=(a1+a6)*6/2=36得a1+a6=12记为1式Sn-S(n-6)=[a(n-5)+an]*6/2=324-144=180得a(n-5)+an=60记为2式又a6+a(n-5)=a1+an