等比数列s=1/q-1什么情况下用 为什么求级数，的等比数列求和，为什么直接是 首项/(1-q)...
\u5df2\u77e5\u6570\u5217S\u662f\u4ee5a1=1,q=-1\u7684\u7b49\u6bd4\u6570\u5217\uff0c\u90a3\u4ed6\u7684Sn\u662f\u591a\u5c11\uff1f\u5982\u4f55\u8bc1\u660e\u51fa\u6765\u662f1/2\uff1fS=1-1+1-1+1-1+1\u2026\u2026\uff0cS=1-(1-1+1-1+1\u2026\u2026)\u7531\u4e8eS=1-1+1-1+1,\u6545S=1-S\uff0c\u6545S=1/2
\u8fd9\u662f\u65e0\u7a77\u9012\u7f29\u7b49\u6bd4\u6570\u5217\u6240\u6709\u9879\u4e4b\u548c\u516c\u5f0f\uff0c \u6210\u7acb\u6761\u4ef6\u662f -1 < q < 1.
\u5426\u5219\u548c\u4e0d\u662f\u56fa\u5b9a\u7684\u6709\u754c\u503c\u3002
\u56e0 S = a1(1-q^n)/(1-q)
\u5f53 -1 q^n = 0.
S = lima1(1-q^n)/(1-q) = a1/(1-q).
绛旓細鏍规嵁鏃犵┓閫掔缉绛夋瘮鏁板垪姹傚拰鍏紡S=a1/(1-q) 鍏朵腑涓╭涓紲1锛屽浜庢湰棰樻湁 a1/(1-q)=1锛屾垨a1=1-q锛涒憼 鍙堬紝鏁板垪鍚勯」鐨勫钩鏂逛緷娆℃槸a1²銆乤1²q²銆乤1²q^4鈥︹1²q^(2n-2)鈥︹︼紝鍏堕椤规槸a1²锛屽叕姣旀槸q²锛屽悇椤圭殑鍜屾槸a1²/(1-q²)=...
绛旓細绛夋瘮鏁板垪鍏紡锛1銆佸畾涔夊紡锛2銆佹眰鍜屽叕寮忥細3銆侀氶」鍏紡锛4銆佷粠绛夋瘮鏁板垪鐨勫畾涔夈侀氶」鍏紡銆佸墠n椤瑰拰鍏紡鍙互鎺ㄥ嚭锛氱瓑宸暟鍒楀叕寮忥細1銆佸畾涔夊紡 瀵逛簬鏁板垪鑻ユ弧瓒筹細鍒欑О璇ユ暟鍒椾负绛夊樊鏁板垪銆傚叾涓紝鍏樊d涓轰竴甯告暟锛宯涓烘鏁存暟銆2銆侀氶」鍏紡 an=a1+(n-1)*d銆傞椤筧1=1锛屽叕宸甦=2銆3銆佸墠n椤瑰拰鍏紡涓猴細Sn=...
绛旓細4銆乹鈮1q鈮1鐨绛夋瘮鏁板垪鐨勫墠2n2n椤癸紝S鍋=a2⋅[1−(q2)n]1−q2S鍋=a2⋅[1−(q2)n]1−q2锛孲濂=a1⋅[1−(q2)n]1−q2S濂=a1⋅[1−(q2)n]1−q2锛屽垯S鍋禨濂=qS鍋禨濂=q銆5銆佺瓑姣旀暟鍒楃殑鍗曡皟鎬э紝鍙栧喅浜庝袱...
绛旓細瑙ｇ瓟锛氭棤绌绛夋瘮鏁板垪{an}鐨勫悇椤瑰拰S鐨勫叕寮忔槸S=a1/(1-q)鈭 a1/(1+1/2)=1/a1 鈭 a1²=3/2 鈭 a1=卤鈭6 /2
绛旓細2銆侀櫎浜嗕娇鐢ㄦ眰鍜屽叕寮忓锛屽彲浠ヤ娇鐢ㄩ掓帹鍏崇郴寮忔潵姹傝В绛夋瘮绾ф暟鐨勫拰銆傞掓帹鍏崇郴寮忔槸锛歋n=a1+q(S(n-1))鍏朵腑锛孲n鏄墠n椤圭殑鍜岋紝a1鏄瓑姣旂骇鏁扮殑绗竴椤癸紝q鏄叕姣斻傝繖涓掓帹鍏崇郴寮忓彲浠ョ敤鏉ラ愰」璁＄畻绛夋瘮绾ф暟鐨勫拰銆傜瓑姣旂骇鏁扮殑姹傚拰杩樺彲浠ヤ娇鐢ㄦ棤绌绛夋瘮鏁板垪鐨勬眰鍜屽叕寮忔潵姹傝В銆3銆佹棤绌风瓑姣旀暟鍒楃殑姹傚拰鍏紡鏄細S=a1/...
绛旓細鍥犱负1+```2鈥︹2^n鏈塶+1 椤 鎵浠 1锛1-2^n+1锛/1-2>127 鏁寸悊鍙緱 2^n+1>126 鎵浠鐨勬渶灏忓间负7锛侊紙2锛夊洜涓哄湪绛夋瘮鏁板垪|an|涓弧瓒硈2=1锛宻4=4 鍏瘮璁句负q 鏍规嵁绛夋瘮鏁板垪鐨勬帹鐞 sk,s2k-sk,s3k-s2k`` 鎴愮瓑姣旀暟鍒楋紝鍏瘮鍙樹负q^k (杩欎釜涓嶆噦鍙互鎺ㄧ悊锛屼竴鑸湁鐨勮佸笀閮戒笉璁诧紝楂樿冨緢灏...
绛旓細鎵浠2S=2^2+2^3+鈥︹+2^(N+1)=S-2+2^(N+1)鎵浠S=2^(N+1)-2 鏂规硶锛绛夋瘮鏁板垪姹傚拰 閲婁箟锛氬鏋滀竴涓暟鍒椾粠绗2椤硅捣锛屾瘡涓椤逛笌瀹冪殑鍓嶄竴椤圭殑姣旂瓑浜庡悓涓涓父鏁帮紝杩欎釜鏁板垪灏卞彨鍋氱瓑姣旀暟鍒椼傝繖涓父鏁板彨鍋氱瓑姣旀暟鍒楃殑鍏瘮锛屽叕姣旈氬父鐢ㄥ瓧姣峲琛ㄧず(q鈮0)銆 娉細q=1 鏃讹紝an涓哄父鏁板垪銆傚嵆a^n=a銆
绛旓細鍜屾槸(3^( n+1)-1锛/2銆傝繖鏄竴涓绛夋瘮鏁板垪锛岃S=1+3+3^2+3^3+鈥︹+3^n 浠ｅ叆绛夋瘮鏁板垪姹傚拰鍏紡Sn=(a1(1-q^n))/(1-q)锛宎1鏄瓑姣旀暟鍒楃殑棣栭」锛宷鏄瓑姣旀暟鍒楃殑姣斿笺係n=(1(1-3^n))/(1-3)=(3^( n+1)-1锛/2銆
绛旓細a(n+2)=(1/3)S(n+1)a(n+2)-a(n+1)=(1/3)[S(n+1)-Sn]=(1/3)a(n+1)a(n+2)/a(n+1)=4/3 鎵浠ワ細{an}鏄叕姣斾负4/3鐨绛夋瘮鏁板垪 an=a1*q^(n-1)=(4/3)^(n-1){a2n}鏄叕姣斾负(4/3)^2=16/9鐨勭瓑姣旀暟鍒楋紝棣栭」涓篴2=4/9 2.a2+a4+a6+鈥︹+a2n =(4/9)[...
绛旓細鍥犱负鏄棤绌烽掔缉绛夋瘮鏁板垪,鎵浠<0,鎵浠^n褰搉瓒嬩簬鏃犵┓澶ф椂q^n瓒嬩簬0,鍗1-q^n瓒嬩簬1,鎵浠ユ棤绌烽掔缉绛夋瘮鏁板垪鐨勬眰鍜屽叕寮忎负S=a1/(1-q).
