一无穷等比数列的第一项是1，公比为q，且1q1<1.若每一项都是它后一项的和的k倍.球
\u82e5\u4e00\u4e2a\u65e0\u7a77\u7b49\u6bd4\u6570\u5217\u7684\u516c\u6bd4q\u6ee1\u8db3|q|<1,\u4e14\u6bcf\u4e00\u9879\u90fd\u7b49\u4e8e\u5b83\u4ee5\u540e\u5404\u9879\u548c\u7684k\u500d,\u5219k\u7684\u53d6a(m)=ka(m+1)/(1-q)=kqa(m)/(1-q)
k=(1-q)/q
=1/q -1
-1<q<0\u62160<q<1
\u6240\u4ee5 1/q1
\u6240\u4ee5 1/q-10
\u5373k0
\u7b2cn\u9879\u662faq^(n-1)
\u4ed6\u7684\u540e\u9762\u662f\u7b49\u6bd4\u6570\u5217\uff0c\u9996\u9879\u662faq^n,\u516c\u6bd4\u662fq
\u6240\u4ee5\u548c=aq^n/(1-q)
\u6240\u4ee5aq^(n-1)/[aq^n/(1-q)]=k
(1-q)q=k
k=q-q^2=-(q-1/2)^2+1/4
-1<q<1, \u4e14 q\u4e0d\u7b49\u4e8e0
\u6240\u4ee5q=1/2,k\u6700\u5927=1/4
q=-1\uff0ck\u6700\u5c0f=-2
q=0\uff0ck=0,k=-q^2+q=0,q\u8fd8\u53ef\u4ee5\u7b49\u4e8e1\uff0c\u4f46\u4e0d\u7b26\u5408|q|<1
\u7efc\u4e0a
-2<k<0,0<k<1/4
求什么?
求n趋于+∞时,Sn的极限吗？
如果是，那么：Sn极限是1/(1-q)
绛旓細1)q>1 a1<a2<a3 鈭碼1=b1,a2=b2,a3=b3 b1=1,q=1+d q²=1+2d=(1+d)² 瑙ｅ緱d=0 涓嶅悎棰樻剰銆2锛0<q<=1 a3<a2<a1 鈭碼1=b3,a2=b2,a3=b1 鍗砨3=1=b1+2d a2=b2=q=b1+d a3=b1=q²d=q-q²q²+2(q-q²)=1 q=1 d=0 涓嶅悎棰...
绛旓細鏃犵┓绛夋瘮鏁板垪鐨鍜屾槸涓涓暟 鍒檤q|<1 鎵浠a-3/2|<1 涓擲=a1/(1-q)S=a,a1=1,q=a-3/2 鎵浠=1/(1-a+3/2)-a^2+5a/2=1 2a^2-5a+2=0 (2a-1)(a-2)=0 |a-3/2|<1 鎵浠=2
绛旓細1 a1=1 a2=q a3=q^2 绛夊樊鏁板垪{bn} b1+b3=2b2 1) 1+q^2=2q (1-q)^2=0 q=1 b1=1 b2= 1 b3=1 鑸 2) 1+q=2q^2 2q^2-q-1=0 q=1鑸 q=-1/2 __b1=-1/2 b2=1/4 b3=1 d=3/4鎴愮珛 3锛塹^2+q=2 q^2+q-2=0 q=1鑸 q=-2 __b1=-2 b2=1 b3=...
绛旓細an=limSn-Sn锛岀‘瀹氫簡n鐨勫硷紝limSn琛ㄧず鏃犵┓鏁板垪鐨勫拰锛屾槸鏃犵┓澶э紝鏁板垪鍓峮椤瑰拰鏄湁闄愭暟锛屽緱鍒扮殑an鏄棤绌峰ぇ锛岃繖涓嶅彲鑳斤紝鑷充簬涓轰粈涔坾q|>=1鏃舵暟鍒楃殑鍜屼负鏃犵┓澶э紝浣犲彧瑕佸鏁板垪鍓峮椤瑰拰鐨勫叕寮忓彇n涓烘棤绌峰ぇ鐨勬瀬闄愶紝寰楀埌鐨勭粨鏋滃氨鏄棤绌峰ぇ銆
绛旓細璁绗竴椤涓篴,鍏瘮涓簈 a/(1-q)=3 a=2,q=1/3 鎵浠ュ墠涓ら」涓2锛2/3=8/3
绛旓細an=a1q^(n-1)=(1/2)^(n-1)Sn=1+1/2+(1/2)^2+鈥︹+锛1/2)^(n-1)Sn/2=1/2+(1/2)^2+鈥︹+(1/2)^(n-1)+(1/2)^n 浜屽紡鐨勪袱杈圭浉鍑忓緱鍒 Sn/2=1-(1/2)^n --->Sn=2-(1/2)^(n-1)褰搉->鏃犵┓澶ф椂锛(1/2)^(n-1)鐨勬瀬闄愭槸0锛屾墍浠->鏃犵┓澶ф椂Sn鐨勬瀬闄...
绛旓細鈭礱1=1,鈭碼n=q^(n-1)鈭-1<q<1 鈭 a(n+1)+a(n+2)+...=a(n+1)/(1-q)=an*q/(1-q)鈭垫瘡涓椤归兘鏄瀹冧互鍚庢墍鏈夐」鍜岀殑m鍊 鈭碼n=man*q/(1-q)m=(1-q)/q=1/q-1 鈭-1<q<0 鈭1/q<-1 鈭1/q-1<-2 鈭磎<-2 ...
绛旓細浠m=a a=kaq(1-q^n)/(1-q)=kaq /(1-q)kaq=a-aq kq=1-q k=1/q -1 k>0 k<-2
绛旓細浠绛夋瘮鏁板垪姹傚拰鍏紡S=a1(1-q^n)/1-q ==>S=(1-1/2^n)/(1-1/2) ==>S=2-2(1/2)^n 鍥犱负(1/2)^n鐨勬瀬闄愪负0 鎵浠鐨勬瀬闄愬氨绛変簬2
绛旓細Sn=(q^n-1)/(q-1)S(n+1)=(q*q^n-1)/(q-1)Sn/S(n+1)=(q^n-1)/(q*q^n-1)鑻<1 limSn/S(n+1)=(0-1)/(0-1)=1 鑻>1 limSn/S(n+1)=(1-1/q^n)/(q-1/q^n)=1/q
