已知数列{an}的前n项和为Sn,且Sn=2an－2(n=1,2,3...),数列{bn}中，b1=1,点(bn,bn+1)在直线x-y+2=0上 已知数列{an}的前n项和为Sn，且Sn=2an-2，（n=...
\u5df2\u77e5\u6570\u5217\uff5ban\uff5d\u7684\u524dn\u9879\u548c\u4e3aSn\uff0c\u4e14Sn=3/2an-1\uff08n\u5c5e\u4e8eN\uff09\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
\uff08 I\uff09\u2235Sn=2an-2\uff0c\u2234\u5f53n\u22652\u65f6\uff0can=Sn-Sn-1=2an-2an-1\u2235an\u22600\uff0c\u2234anan?1\uff1d2\uff08n\u22652\uff09\uff0c\u5373\u6570\u5217{an}\u662f\u7b49\u6bd4\u6570\u5217\uff0e\u2235Sn=2an-2\uff0c\u2234\u5f53n=1\u65f6\uff0ca1=2\uff0c\u2234an\uff1d2n \u2026\uff083\u5206\uff09\u2235\u70b9P\uff08bn\uff0cbn+1\uff09\u5728\u76f4\u7ebfx-y+2=0\u4e0a\u2234bn-bn+1+2=0\uff0c\u2234bn+1-bn=2\u5373\u6570\u5217{bn}\u662f\u7b49\u5dee\u6570\u5217\uff0c\u53c8b1=1\uff0c\u2234bn=2n-1 \u2026\uff086\u5206\uff09\uff08 II\uff09Sn=a1b1+a2b2+\u2026+anbn=1\u00d72+3\u00d722+\u2026+\uff082n-1\uff09\u00d72n \u2460\uff087\u5206\uff09\u22342Sn=1\u00d722+3\u00d723+\u2026+\uff082n-1\uff09\u00d72n+1\u2461\u2460-\u2461\u5f97\uff1a-Sn=1\u00d72+2\uff0822+23+\u2026+2n\uff09-\uff082n-1\uff09\u00d72n+1\u2461\u2026\uff089\u5206\uff09\u2234Sn\uff1d(2n?3)?2n+1+6\uff0810\u5206\uff09\u2235Sn\uff1c167\uff0c\u5373\uff082n-3\uff09?2n+1+6\uff1c167\u4e8e\u662f\uff082n-3\uff09?2n+1\uff1c161\uff0811\u5206\uff09\u53c8\u7531\u4e8e\u5f53n=4\u65f6\uff0c\uff082n-3\uff09?2n+1=160\u5f53n=5\u65f6\uff0c\uff082n-3\uff09?2n+1=448\uff0813\u5206\uff09\u6545\u6ee1\u8db3\u6761\u4ef6Sn\uff1c167\u6700\u5927\u7684\u6b63\u6574\u6570n\u4e3a4\uff0814\u5206\uff09
解：（1）由Sn = 2an—2 可得，当n=1时，S1 = a1 = 2 a1—2解得a1 = 2 又Sn-1 = 2an-1—2
则Sn — Sn-1 = an = 2an—2—（2an-1—2）=2an—2an-1
整理可得，an = 2 an-1 ，为等比数列，公比为q = 2
故an = a1•qn-1 = 2•2n-1 = 2n ，n∈N+
因为点（bn ，bn+1）在直线 x—y+2=0上，
则有bn—bn+1+2=0 ，即bn+1—bn=2
此数列为等差数列，公差为d = 2 ，又b1 = 1
故bn=b1+(n—1) d = 1+(n—1)•2 = 2 n—1
当n=1时，b1 = 1 则bn = 2 n—1 , n∈N+
(2) 由（1）可知an•bn = 2n•(2 n—1)= 2n+1•n—2n
所以：
Sn = 22•1—2+23•2—22 +24•3—23+…+2n•(n—1) —2n-1+2n+1•n—2n
=23•1+24•2+…+2n•(n—2) +2n+1•n—2 ①
2Sn=24•1+25•2+…+2n+1•(n—2) +2n+2•n—4 ②
①式—②式，得
—Sn = 23+24+25+…+2n +2n+1•2—2n+2•n+2
=2+(23—23•2n-2)／(1—2) +2n+2•(1—n)
=—6—2n+1•(2 n—3)
综上，Sn =6+2n+1•(2 n—3) 解毕.
(在文档上解完粘贴上来效果不一样，楼主如若看不清楚，可留下邮箱，我可以将文档发给你)
(1)由Sn=2an-2得：
a1=s1=2a1-2
a1=2;
an=S(n)-S(n-1)=2an-2-(a(n-1)-2)
得an/a(n-1)=2,{an}是等比数列
an=2^n
bn-bn+1+2=0得b(n+1)-bn=2
得{bn}是等差数列，bn=2n-1
(2)Sn=2*1+2^2*(2*2-1)+……+2^n(2*n-1)
2Sn= 2^2*1+2^3*(2*2-1)+……+2^n(2*n-3)+2^(n+1)(2*n-1)
-Sn=2+2^2*(2*2-2)+……+2^n*2-2^(n+1)(2*n-1)
Sn=2-2^3-2^4-……+2^(n+2)*(n-1)
=2-8(2^(n-2)-1)+2^(n+2)*(n-1)
an=二分之一的（n－2）次方，bn＝3－2n
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...
