{an}为无穷等比数列,{an}中每一项都是它后面所有项和的k倍,且a1=1,求k的取值范围
设公比是q由题意知，|q|<1, an=ka(n+1)/(1-q) 则1-q=ka(n+1)/an=kq 解得 q=1/(1+k)因为，|q|<1 ，所以 |1/(1+k)|<1 解得k0
绛旓細鎴戣涓鸿繖鏄厖瑕佹潯浠躲{an}锛屾棤绌风瓑姣旀暟鍒梴an}锛鎵浠ラ椤筧1锛屽叕姣攓閮戒笉涓0 1# lim an =lim aq^(n-1)=0 鈫 0<|q| <1 lim (Sn) = lim (a1(1-q^(n-1))/(1-q)) = a1/(1-q) 瀛樺湪 2# 鑻 lim (Sn) 瀛樺湪 1& 褰 q =1 鏃讹紝 Sn = na1, lim(Sn) 涓嶅瓨鍦 2& ...
绛旓細PS:鎴戠敤a[n]琛ㄧず绗琻椤.s[n]琛ㄧず鍓峮椤瑰拰.绗1涓鐩 瑙:褰撳叕姣攓=1鏃,s[n]=a[1]+a[2]+...+a[n]=na[1]鍏朵腑a[1]涓轰竴甯告暟,姝ゆ椂lim(s[n])涓衡垶 褰撳叕姣攓=-1鏃,鑻涓哄鏁,s[n]=a[1]+a[2]+...+a[n]=a[1],鑻涓哄伓鏁,s[n]=0 鑻q|>1,lims[n]=鈭 缁间笂鎵杩皘...
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...
绛旓細鐢遍鎰忓彲寰楋細a11?q锛2锛寍q|锛1涓攓鈮0锛屸埓a1=2锛1-q锛夛紝鈭0锛渁1锛4涓攁1鈮2锛屽垯棣栭」a1鐨勫彇鍊艰寖鍥存槸锛0锛2锛夆埅锛2锛4锛夛紟鏁呯瓟妗堜负锛氾紙0锛2锛夆埅锛2锛4锛
绛旓細鐢遍鎰忕煡锛歛1=a1q+路路路+anq+路路路=a1q锛1-q^n锛/(1-q)锛坬瓒嬩簬鏃犵┓澶э級锛屾樉鐒舵湁|q|<1锛屾墍浠ユ湁q=1/2 鐢盿3=3寰楀埌a2=6锛屾墍浠ユ墍鏈夊伓鏁伴」鐨勫拰涓6+6*1/4+路路路+6*1/4^n+路路路=8
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=...
绛旓細鈭礢n=a1(1-qn)1-q锛0锛渱q|锛1锛屸埓limn鈫掆垶Sn=a11-q=1a1锛屸埓q=1-a12锛屸埖a1鈭(0锛22)锛屸埓q鈭(12锛1)锛庡洜姝ゅ叕姣攓鐨勫彇鍊艰寖鍥存槸(12锛1)锛庢晠绛旀涓(12锛1)锛
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
