已知数列{An}的前n项和为sn，3sn=an-1(n属于整数).(1)求证：数列{an}是等比数列；(2)求an和sn为多少
\u5df2\u77e5\u6570\u5217{An}\u7684\u524dn\u9879\u548c\u4e3asn\uff0c3sn=an-1(n\u5c5e\u4e8e\u6574\u6570).(1)\u6c42\u8bc1\uff1a\u6570\u5217{an}\u662f\u7b49\u6bd4\u6570\u5217\uff1b(2)\u6c42an\u548csn\u4e3a\u591a\u5c11(1)\u8bc1\u660e\uff1a\u22353sn=an-1
\u22343S( n+1)=a(n+1)-1
\u4e24\u5f0f\u76f8\u51cf\uff1a
3S(n+1)-3Sn=a(n+1)-an
\u53c83S(n+1)-3Sn=3[S(n+1)-Sn]=3a(n+1)
\u22343a(n+1)=a(n+1)-an
\u22342a(n+1)=-an
\u2234a(n+1)/an=-1/2
\u2234\u6570\u5217{an}\u662f\u7b49\u6bd4\u6570\u5217
\uff082\uff09\u89e3\uff1a\u2235{an}\u662f\u7b49\u6bd4\u6570\u5217,\u516c\u6bd4\u4e3a-1/2
\u53c83a1=3S1=a1-1
\u2234a1=-1/2
\u2234an=(-1/2)*(-1/2)^(n-1)=(-1/2)ⁿ
Sn=(-1/2)*[1-(-1/2)^n]/(1+1/2)
=(-1/3)*[1-(-1/2)ⁿ]
\u2474Sn=3/2an-1\uff0c\u2234S(n-1)=3/2A(n-1)-1,\u4e24\u5f0f\u76f8\u51cf\u6574\u7406\u5f97\uff1a
An/A(n-1)=3,{an}\u662f\u7b49\u6bd4\u6570\u5217\uff0c\u516c\u6bd4\u4e3a3\uff0c\u9996\u9879\u7531Sn=3/2an-1\u5f97\uff0c\u53e6n=1,S1=a1
\u5f97\uff1aA1=2\uff0c\u2234An=2*3^(n-1)
\u2475B(n+1)-Bn=2*3^(n-1)
\u2236Bn=(Bn-B(n-1))+(B(n-1)-B(n-2))+....+(B2-B1)+B1,\u8fd9\u662f\u8fed\u4ee3\u6cd5\uff0c\u7528\u5927\u5199\u5b57\u6bcd\u4fbf\u4e8e\u533a\u522b\u4e0b\u6807
=2*3^(n-2)+2*3^(n-3)+...+2*3^0+5
=2(3^(n-2)+3^(n-3)+...+3^0)+5
=2*(1-3^(n-1))/(1-3)+5
=3^(n-1)+4
∵3sn=an-1
∴3S( n+1)=a(n+1)-1
两式相减：
3S(n+1)-3Sn=a(n+1)-an
又3S(n+1)-3Sn=3[S(n+1)-Sn]=3a(n+1)
∴3a(n+1)=a(n+1)-an
∴2a(n+1)=-an
∴a(n+1)/an=-1/2 （定值）
∴数列{an}是等比数列
（2）
∵{an}是等比数列,公比为-1/2
又3a1=3S1=a1-1
∴a1=-1/2
∴an=(-1/2)*(-1/2)^(n-1)=(-1/2)ⁿ
Sn=(-1/2)*[1-(-1/2)^n]/(1+1/2)
=(-1/3)*[1-(-1/2)ⁿ]
(1)
∵3sn=an-1
∴3S( n+1)=a(n+1)-1
两式相减：
3S(n+1)-3Sn=a(n+1)-an
又3S(n+1)-3Sn=3[S(n+1)-Sn]=3a(n+1)
∴3a(n+1)=a(n+1)-an
∴2a(n+1)=-an
∴a(n+1)/an=-1/2 （定值）
∴数列{an}是等比数列
（2）
∵{an}是等比数列,公比为-1/2
又3a1=3S1=a1-1
∴a1=-1/2
∴an=(-1/2)*(-1/2)^(n-1)=(-1/2)ⁿ
Sn=(-1/2)*[1-(-1/2)^n]/(1+1/2)
=(-1/3)*[1-(-1/2)ⁿ]
3sn=an-1 (1)
3s(n-1)=a(n-1)-1 (2)
(1)-(2):
3an=an-a(n-1)
2an=-a(n-1)
an/a(n-1)=-1/2
且3s1=3a1=a1-1
a1=-1/2
即{An}是以为-1/2首项-1/2为公比的等比数列
an=-1/2×(-1/2)^(n-1)
=(-1/2)^n
sn=(-1/2)*(1-(-1/2)^n)/(1+1/2)
=[(-1/2)^n-1]/3
希望能帮到您，欢迎追问
1、有题意可知：3(sn-1)=(an-1)-1(n大于1）,3sn-3(san-1)=an-1-(an-1-1),即3an=an-(an-1),得2an=-(an-1),an/（an-1)=-1/2即公比q=-1/2，当n=1时，由题可知：3a1=a1-1，即a1=-1/2，首项a1不为0，公比不为0，所以数列an为等比数列。
2、和公式sn==a1(1-q^n)/(1-q) 即：sn=-1/2（1-（-1/2）^n）（1+1/2），得：sn=-3/4（1-（-1/2）^n）结束
（1）3Sn=an-1 ① 3S(n-1)=a(n-1)-1② ②-①得an/(an-1)=-1/2所以an是等比数列
（2）3a1=a1-1 a1=-1/2将q=-1/2代入得an=(-1/2)^n sn=1/3×（-1/2)^n-1/3
绛旓細瑙ｏ細锛1锛夌敱Sn = 2an鈥2 鍙緱锛屽綋n=1鏃讹紝S1 = a1 = 2 a1鈥2 瑙ｅ緱a1 = 2 鍙圫n-1 = 2an-1鈥2 鍒橲n 鈥 Sn-1 = an = 2an鈥2鈥旓紙2an-1鈥2锛=2an鈥2an-1 鏁寸悊鍙緱锛宎n = 2 an-1 锛屼负绛夋瘮鏁板垪锛屽叕姣斾负q = 2 鏁卆n = a1•qn-1 = 2•2n-1 = 2n ...
绛旓細(1)鈭3sn=an-1 鈭3S( n+1)=a(n+1)-1 涓ゅ紡鐩稿噺锛3S(n+1)-3Sn=a(n+1)-an 鍙3S(n+1)-3Sn=3[S(n+1)-Sn]=3a(n+1)鈭3a(n+1)=a(n+1)-an 鈭2a(n+1)=-an 鈭碼(n+1)/an=-1/2 锛堝畾鍊硷級鈭鏁板垪{an}鏄绛夋瘮鏁板垪 锛2锛夆埖{an}鏄瓑姣旀暟鍒,鍏瘮涓-1/2 鍙3a1=3...
绛旓細鍥犱负鏍规嵁S(n) - S(n-1)姹俛(n)鐨勫叕寮忕殑鏃跺欙紝瑕佷繚璇丼(n-1)鏈夋剰涔夛紝鍗宠淇濊瘉n-1>=1,鍗硁>=2锛屾墍浠ヨ繖涓姝ユ眰鍒扮殑鏄痭>=2鐨勬椂鍊欑殑鍏紡銆傛墍浠 a1 瑕佸彟澶栨眰锛屽叾瀹炲緢绠鍗曪紝鍙灏 n=1甯﹀叆Sn姹傚緱a(1)銆傜畻鍒拌繖閲岃娉ㄦ剰锛屾暟瀛﹁绌剁畝娲佺編锛岃繖閲岃鎶妌=1甯﹀叆鍓嶉潰姹傚緱鐨an鍏紡涓湅鏄惁婊¤冻锛...
绛旓細a3 = 3a1锛岋紙2锛夊皢杩欎竴琛ㄨ揪寮忎唬鍏ワ紙1锛夛紝寰楀埌锛歛5 = [ 42 - 17a1] / 5. 锛3锛夊啀鍒╃敤绛夊樊鏁板垪閫氶」鍏紡锛岃鍏樊涓篸锛屽垯鏈塧3 = a1 + 2d, a5 = a1 + 4d锛岄兘浠ｅ叆鍒帮紙2锛夊紡鍜岋紙3锛夊紡涓紝灏卞彲浠ヨВ鍑猴細a1 = d = 1.浜庢槸Sn = (a1 + an)n / 2 = [1 + (1 + (n-1)*...
绛旓細=s(n)-s(n+1)/s(n)s(n+1)=2a(n+1)/a(n+1)=2 鍥犱负a1=s1=1/2 鎵浠2=-2s2s1 =-1(1/2+a2)鎵浠2=-1/4 s2=1/4 鎵浠1=2 b2=4 鎵浠ュ綋n>=2鏃禸(n)=2n 鍙堝洜涓篵1=2 鎵浠bn}鏄浠2涓洪椤癸紝鍏樊涓2鐨勭瓑宸鏁板垪 鎵浠n=2n p(n)=n(n+1)ps:鏍煎紡鍙兘鏈夌偣閿欙紝...
绛旓細Sn=2an-2 閭ｄ箞 S(n+1)=2a(n+1)-2 涓ゅ紡鐩稿噺 S(n+1)-Sn=2a(n+1)-2an a(n+1)=2a(n+1)-2an a(n+1)=2an 鎵浠an鏄浠2涓哄叕姣旂殑绛夋瘮鏁板垪 褰搉=1鏃 s1=2a1-2=a1 鎵浠1=2 鎵浠n=2^n
绛旓細绗竴闂細a1=1,S1=1 {Sn/n}鐨绗竴椤筍1/1 涓1/1=1 鍒橲n/n锛1+锛坣锛1锛壝1锛漬 Sn锛漬^2 S(n-1)=(n-1)^2 an=Sn-S(n-1)=n^2-(n-1)^2锛2n锛1 绗簩闂細鍥犱负an锛峚(n-1)=[2n锛1]-[2(n-1)锛1]=2 鎵浠an涓鍏樊涓2鐨勭瓑宸鏁板垪銆傚綋K涓哄鏁版椂 锛 (-1...
绛旓細An = Sn-S(n-1)= an^2+bn+c-a(n-1)^2-b(n-1)-c = a(n+n-1)(n-n+1)+b(n-n+1)= a(2n-1)+b 鎵浠 A(n-1) = a[2(n-1)-1]+b = a(2n-3)+b 鎵浠 An-A(n-1) = a(2n-1)+b-a(2n-3)-b = 2a 鏄父鏁 鎵浠鏁板垪{An}浠庣浜岄」璧凤紝鏄叕宸负2a鐨勭瓑宸...
绛旓細Sn=nan,S(n-1)=(n-1)a(n-1),鐩稿噺锛歛n=nan - (n-1)a(n-1),鈭(n-1)an=(n-1)a(n-1),an=a(n-1)锛屽嵆an=1
绛旓細a(n-1)=(n-2)a(n-2)/n 鈥︹︹2=a1/3 杩炰箻 a2a3...an=a1a2...a(n-1)[(n-1)(n-2)...1]/[(n+1)n...3]=2a1a2...a(n-1)/[n(n+1)]an=2a1/[n(n+1)]=2/[n(n+1)]n=1鏃讹紝a1=2/(1脳2)=1锛屽悓鏍锋弧瓒炽鏁板垪{an}鐨閫氶」鍏紡涓篴n=2/[n(n+1)]绗簩...
