{an}为无穷等比数列,公比|q|<1,且每一项与它以后的所有项的和之比2:3,求公比q
\u65e0\u7a77\u7b49\u6bd4\u6570\u5217{an}\u4e2d\uff0c\u516c\u6bd4\u4e3aq\uff0c\u4e14\u6240\u6709\u9879\u7684\u548c\u4e3a2/3\uff0c\u5219a1\u7684\u8303\u56f4\u662f________\u8be6\u7ec6\u6c42\u89e3\u8fc7\u7a0b\u5982\u4e0b
an=a1\u00d7q^(n-1)=9/8\u00d7(2/3)^(n-1)=1/3
\u6240\u4ee5 (2/3)^(n-1)=(1/3)\u00d7(8/9)
(2/3)^(n-1)=8/27
\u5f97 n-1=3
\u6240\u4ee5 n=4
an/Rn=an/[a(n+1)/(1-q)]=(1-q)/q=2/3;
解得q=3/5.
绛旓細鍙緱锛an^2涔熸槸涓涓瓑姣旀暟鍒楋紱棣栭」涓篴1^2=1/4锛鍏瘮涓猴細q^2=1/4;鎵浠ワ細(a2n)^2=[(1/2)^2n]^2 [a2(n-1)]=[(1/2)^2(n-1)](a2n)^2/[a2(n-1)]^2=(1/2)^4 鎵浠n涔熸槸绛夋瘮鏁板垪锛棣栭」涓篴2^2=1/16; 鍏瘮锛歲^4=1/16; 鏄棤绌风瓑姣旀暟鍒 鏍规嵁鏃犵┓绛夋瘮鏁板垪鐨勬ц川Sn=a1...
绛旓細锛堟湰灏忛婊″垎13鍒嗭級锛堚厾锛夎В锛氣埖绛夋瘮鏁板垪{an}涓紝a1=4锛宷=12锛屸埓a1=4锛宎2=2锛宎3=1锛屼笖褰搉锛3鏃讹紝0锛渁n锛1锛庘︼紙1鍒嗭級鈭礲n=[an]锛屸埓b1=4锛宐2=2锛宐3=1锛屼笖褰搉锛3鏃讹紝bn=[an]=0锛庘︼紙2鍒嗭級鈭碩n=4锛宯=16锛宯=27锛宯鈮3锛庘︼紙3鍒嗭級锛堚叀锛夎瘉鏄庯細鈭礣n=2n+1锛坣鈮2014锛夛紝...
绛旓細鈭礱1=1,鈭an=q^(n-1)鈭-1<q<1 鈭 a(n+1)+a(n+2)+...=a(n+1)/(1-q)=an*q/(1-q)鈭垫瘡涓椤归兘鏄畠浠ュ悗鎵鏈夐」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Sn=a1(1-q^n)/(1-q)鐢变簬limSn=lim[a1(1-q^n)/(1-q)]=1,鏁呭繀鏈墊q|<1锛屼笖limSn=a1/(1-q)=1 q=1-a1 鏁呮湁锛殀q|=|1-a1|<1 0<a1<2
绛旓細瑙o細an鐨勫墠涓夐」涓 a1=1,a2=q,a3=q²bn鐨勫墠涓夐」涓 b1,b2=b1+d,b3=b1+2d,涓攂1<b2<b3 1銆佹眰q鍊 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=...
绛旓細娌℃湁鍒嗗鍔憋紝鎵浠ユ湁鍏磋叮鍥炵瓟鐨勪汉澶皯銆係1=a1(1-q^n)/(1-q)鐢变簬limSn=lim[a1(1-q^n)/(1-q)]=1,鏁呭繀鏈墊q|<1锛屼笖limSn=a1/(1-q)=1 q=1-a1 鏁呮湁锛殀q|=|1-a1|<1 0<a1<2
绛旓細鍗砈n/Sn+1=(1-q^n+1)/(1-q^n)锛涙鏃跺垎涓ょ鎯呭喌璁ㄨ锛氣憼 0<q<1鏃讹紝lim(q^n)=0锛岀浉瀵逛簬1锛宷^n鍙互蹇界暐涓嶈锛屽嵆lim(Sn/Sn+1)=lim銆(1-q^n+1)/(1-q^n)銆=(1-0)/(1-0)=1;鈶 q>1鏃讹紝lim(q^n)瓒嬪悜浜鏃犵┓澶э紝Sn/Sn+1=(1-q^n+1)/(1-q^n)=(q^n+1 - 1)...
绛旓細an=q^(n-1)sn=(q^n-1)/(q-1)limsn=1/(1-q)an涔嬪悗鎵鏈夐」涔嬪拰Tn Tn=1/(1-q)-sn =1/(1-q)-(q^n-1)/(q-1)=-q^n/(q-1)鎵浠 an=-kq^n/(q-1)kq^n/(q-1)=q^(n-1)qk=q-1 q=1/(1-k)鈮0 -1锛渜=1/(1-k)锛1 -1锛1/(1-k)锛0鎴0锛1/(1-k)锛...
绛旓細Sn=a1(1-q^n)/(1-q)鐢变簬limSn=lim[a1(1-q^n)/(1-q)]=2,鏁呭繀鏈墊q|<1锛屼笖limSn=a1/(1-q)=2 a1=(1-q)*2 q=1-a1/2 鍙坬<0,鍒-1<q<0 鏁呮湁锛-1<1-a1/2<0 -2<-a1/2<-1 2<a1<4.
绛旓細瑙g瓟锛鏃犵┓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