数列an中a1=1,{an/n}是以2为公比的等比数列（1）求{an}的通项公式（2）求前n项和Sn 求详尽过程
\u5df2\u77e5\u7b49\u6bd4\u6570\u5217{an}\u7684\u524dn\u9879\u548c\u4e3aSn=a2^n+b,\u4e14a1=3,\u6c42a,b\u7684\u503c\u53ca\u6570\u5217{an}\u7684\u901a\u9879\u516c\u5f0fSn=a2^n+b
a1=2a+b=3
a1+a2=4a+b,a2=2a
a1+a2+a3=8a+b.a3=4a
\u7b49\u6bd4\u6570\u5217\uff0ca2/a1=a3/a2
a=3,b=-3
an=3*2^(n-1)
\u7528\u9519\u4f4d\u76f8\u51cf\u6cd5\uff1a
bn=(2/3)*n/2^n
Tn=(2/3)*[1/2+2/2^2+...+n/2^n]
Tn/2=(2/3)*[[1/2^2+2/2^3+...+(n-1)/2^n+n/2^(n+1)]
\u76f8\u51cf\uff1aTn/2=\uff082/3\uff09*[1/2+1/2^2+...+1/2^n-n/2^(n+1)]
Tn=(4/3)*[1-(2+n)/2^(n+1)]
\u9898\u76ee\u8981\u6c42\u7684\u662fan\u7684\u901a\u9879\u516c\u5f0f\uff0c\u4e0d\u662fSn
解：(1)
设bn=an/n，b1=1，q=2
∴bn=2^(n-1)
∴an=n2^(n-1)
(2)
Sn=a1+a2+a3+……+a(n-1)+an
=1×2^0+2×2^1+3×2^2+……+(n-1)×2^(n-2)+n×2^(n-1)
2Sn =1×2^1+2×2^2+3×2^3+……+(n-1)×2^(n-1)+n×2^n
2Sn-Sn=-1×2^0+(1-2)×2^1+(2-3)×2^2+……+[n-2-(n-1)]×2^(n-2)+(n-1-n)×2^(n-1)+n×2^n
Sn=-1-2^1-2^2-2^3-2^4-……-2^(n-2)-2^(n-1)+n×2^n
=n2^n-1-(2^1+2^2+2^3+……+2^n)
∵2^1+2^2+2^3+……+2^n=2^(n+1)-2
∴ Sn=n2^n-1-2^(n+1)+2
=n2^n-2^(n+1)+1
或=(n-2)2^n+1
解:(1)由题可知an=a1×q^(n-1)=1×2^(n-1)=2^(n-1);(2)由(1)可知Sn=a1(1-q^n)/(1-q)=1×(1-2^n)/(1-2)=2^n-1。
绛旓細a1/1=1/1=1 鏁板垪{an/n}鏄互1涓洪椤癸紝2涓哄叕宸殑绛夊樊鏁板垪銆傚張bn=an/n 鏁板垪{bn}鏄互1涓洪椤癸紝2涓哄叕宸殑绛夊樊鏁板垪銆俠n=1+2(n-1)=2n-1 鏁板垪{bn}鐨勯氶」鍏紡涓篵n=2n-1銆(2)an/n =2n-1 an=n(2n-1)=2n²-n n=1鏃讹紝a1=2-1=1锛鍚屾牱婊¤冻 鏁板垪{an}鐨勯氶」鍏紡涓篴n=2n&...
绛旓細a(n+1)-(3/10)*4^(n+1)=-[an-(3/10)4^n]鎵浠an-(3/10)*4^n}鏄叕姣斾负-1鐨勭瓑姣鏁板垪 棣栭」a1-(3/10)*4==1-6/5=-1/5 鏁卆n-(3/10)*4^n=(-1/5)*(-1)^(n-1)鎵浠ラ氶」鍏紡涓 an=(3/10)*4^n-(1/5)*(-1)^(n-1)
绛旓細鎵浠4n²+2鈮²-n+6 閭ｄ箞n²-n+6鈮2脳3^n 鎵浠(n²-n+6)/[2脳3^n]鈮1 鎵浠1/[3^(n-1)]鈮an鈮1
绛旓細宸茬煡鏁板垪{an}涓,a1=1,an+1=(n/n+1)an,姹an鐨閫氬悜鍏紡,鐢ㄥ彔鍔犳硶 娉曚竴锛氭瀯閫犵瓑姣旀垨绛夊樊鏁板垪銆 a(n+1)=nan/(n+1) (n+1)a(n+1)=nan锛1脳a1=1. 鈭存暟鍒梴nan}鏄椤逛负1锛屽叕姣斾负1鐨勭瓑姣旀暟鍒椼 鎴栨暟鍒梴nan}鏄椤逛负1锛屽叕宸负0鐨勭瓑宸暟鍒椼 nan=1脳a1=1锛鏁卆n=1/n銆
绛旓細瑙ｏ細a(n+1)=[n/(n+1)]an (n+1)a(n+1)=nan 1路a1=1路1=1 鏁板垪{nan}鏄悇椤瑰潎涓1鐨勫父鏁鏁板垪 nan=1 an=1/n n=1鏃讹紝a1=1/1=1锛宎1=1鍚屾牱婊¤冻琛ㄨ揪寮 鏁板垪{an}鐨勯氶」鍏紡涓篴n=1/n
绛旓細鍒╃敤锛a1=1鍙奱(n锛1)=2an/(2锛媋n)锛屽緱锛歛1=1 a2=2/3 a3=1/2=2/4 a4=2/5 鐚滄祴锛歛n=2/(n锛1)璇佹槑锛1銆佸綋n=1鏃讹紝an=a1=2/(1锛1)=1锛婊¤冻锛2銆佽锛氬綋n=k鏃讹紝ak=2/(k锛1)鍒欏綋n=k锛1鏃讹紝a(k锛1)=2ak/(2锛媋k)銆愪互ak=2/(k锛1)浠ｅ叆銆=2/[(k锛1)锛1]鍗...
绛旓細锛1锛夌敱a1=3锛宎n+1+an=3•2n锛宯鈭圢*锛庡緱锛歛n+1−2n+1锛−(an−2n)锛屾墍浠鏁板垪{an−2n}鏄互a1-2=1涓洪椤癸紝鍏瘮涓-1鐨勭瓑姣旀暟鍒楋紝鈭碼n−2n=锛-1锛塶-1锛屾墍浠n锛2n+(−1)n−1锛涳紙2锛夊亣璁惧瓨鍦ㄨ繛缁笁椤筧n-1锛宎n锛宎n+1鎴愮瓑宸暟鍒...
绛旓細鈭存柊鏁板垪{an/2^(n-1)}灏辨垚浜嗕竴涓互a1/2^0=1涓洪椤 1涓哄叕宸殑绛夊樊鏁板垪 鈭碼n=n脳2^(n-1)鈭碨n=a1+a2+...+an =1.2^0+2.2^1+...+n.2^(n-1) (1)鈭2Sn= 1.2^1+2.2^2+...+锛坣-1).2^(n-1)+n.2^n 锛2锛夛紙1锛-锛2锛夊緱锛-Sn=2^0+2^1+2^2+.....
绛旓細an=(3n-2)路2ⁿ⁻¹n=1鏃讹紝a1=(3路1-2)路2⁰=1锛宎1=1鍚屾牱婊¤冻琛ㄨ揪寮 鏁板垪{an}鐨勯氶」鍏紡涓篴n=(3n-2)路2ⁿ⁻¹瑙ｉ鎬濊矾锛氭绫绘眰鏁板垪閫氶」鍏紡鐨勯鐩紝鍏抽敭鏄眰寰楀悗椤逛笌鍓嶉」涔嬮棿鐨勫叧绯诲紡锛屼緥濡傛湰棰橈紝鍗虫眰寰梐(n+1)涓an鐨鍏崇郴寮忋傝繖鏍峰彲浠...
绛旓細鍒嗘瀽锛氱敱a1=1锛宎n+1=an2-1锛屾妸n=1锛2锛3锛4浠ｅ叆鍙垎鍒眰瑙2锛宎3锛宎4锛宎5锛屼粠鑰屽彲姹 瑙ｇ瓟锛氳В锛氣埖a1=1锛宎n+1=an2-1 鈭碼2=a12-1=0 a3=a22-1=-1 a4=a32-1=0 a5=a42-1=-1 a1+a2+a3+a4+a5=1+0-1+0-1=-1 鏁呯瓟妗堜负锛-1 鐐硅瘎锛氭湰棰樹富瑕佽冩煡浜嗙敱鏁板垪鐨閫掓帹鍏紡姹傝В...
