已知数列{an}的首项a1=1,且满足an+1=an/2an+1(n∈N*) 已知正项数列{an}满足：a1=1，且（n+1）an+12=...
\u5df2\u77e5\u6570\u5217{an}\u4e2d\uff0ca1=1\uff0ca2=2\uff0can+1=2an+3an-1\uff08n\u22652\u4e14n\u2208N*\uff09\uff0e\uff08I\uff09\u8bc1\u660e\u6570\u5217{an+an+1}\u662f\u7b49\u6bd4\u6570\u5217\uff1b\uff08II\uff08I\uff09\u8bc1\u660e\uff1a\u56e0\u4e3aan+1=2an+3an-1\uff0c\u6240\u4ee5an+1+an=3\uff08an+an-1\uff09\uff0c\u6240\u4ee5an+1+anan+an?1=3\u662f\u5e38\u6570\uff0c\u6240\u4ee5\u6570\u5217{an+an+1}\u662f\u4ee5a1+a2=3\u4e3a\u9996\u9879\uff0c\u7b49\u6bd4\u4e3a3\u7684\u7b49\u6bd4\u6570\u5217\uff1b\uff08II\uff09\u7531\uff08\u2160\uff09\u5f97an+1+an=3n\uff0c\u2026\u2460\uff0c\u53c8an+1=2an+3an-1\uff08n\u22652\u4e14n\u2208N*\uff09\uff0e\u5f97an+1-3an=-\uff08an-3an-1\uff09\uff0c\uff08n\u22652\u4e14n\u2208N*\uff09\uff0e\u5373an+1?3anan?3an?1=-1\uff0c\u5e38\u6570\uff0c\u6240\u4ee5\u6570\u5217{an+1-3an}\u662f\u4ee5-1\u4e3a\u9996\u9879\uff0c\u516c\u6bd4\u4e3a-1\u7684\u7b49\u6bd4\u6570\u5217\uff0can+1-3an=\uff08-1\uff09n\uff0c\u2026\u2461\uff0c\u89e3\u2460\u2461\u5f97\uff0can=14?3n?14?(?1)n\uff0c\u2234a1+a2+\u2026an=14\uff0831+32+33+\u2026+3n\uff09-14[\uff08-1\uff09+\uff08-1\uff092+\uff08-1\uff093+\u2026+\uff08-1\uff09n]=18[3n+1+(?1)n+1?2] \uff08n\u2208N*\uff09\uff0e
\uff08I\uff09\u2235\uff08n+1\uff09an+12-nan2+an+1an=0\u2234an+1\uff1d?1\u00b11+4n(n+1)2(n+1)an\uff1dnn+1an\uff08\u53e6\u89e3-an\u4e0d\u5408\u9898\u610f\u820d\u53bb\uff09\uff0c\u2234a2a1?a3a2anan?1\uff1d12\uff0c\u5373 ana1\uff1d1n\uff0can\uff1d1n\uff0cn\u2208N+\uff0c\uff08II\uff09\u7531\uff08I\uff09\u5f97\uff1aTn=n!\uff0c\u5f53x\uff1e0\u65f6\uff0cTn\uff1exnex\u7b49\u4ef7\u4e8exn\uff1cn!ex \u2460\u4ee5\u4e0b\u7528\u6570\u5b66\u5f52\u7eb3\u6cd5\u8bc1\u660e\uff1a\u2460\u5f53n=1\u65f6\uff0c\u8981\u8bc1x\uff1cex\uff0c\u4ee4g\uff08x\uff09=ex-x\uff0c\u5219g\u2032\uff08x\uff09=ex-1\uff1e0\uff0c\u2234g\uff08x\uff09\uff1eg\uff080\uff09=1\uff1e0\uff0c\u5373x\uff1cex \u6210\u7acb\uff1b\u2461\u5047\u8bbe\u5f53n=k\u65f6\uff0c\u2460\u5f0f\u6210\u7acb\uff0c\u5373xk\uff1ck!ex\uff0c\u90a3\u4e48\u5f53n=k+1\u65f6\uff0c\u8981\u8bc1xk+1\uff1c\uff08k+1\uff09!ex\u4e5f\u6210\u7acb\uff0c\u4ee4h\uff08x\uff09=\uff08k+1\uff09!ex-xk+1\uff0c\u5219h\u2032\uff08x\uff09=\uff08k+1\uff09!ex-\uff08\uff08k+1\uff09xk=\uff08k+1\uff09\uff08k!ex-xk\uff09\uff0c\u7531\u5f52\u7eb3\u5047\u8bbe\u5f97\uff1ah\u2032\uff08x\uff09\uff1e0\uff0c\u2234h\uff08x\uff09\uff1eh\uff080\uff09=\uff08k+1\uff09!\uff1e0\uff0c\u5373xk+1\uff1c\uff08k+1\uff09!ex\u4e5f\u6210\u7acb\uff0c\u7531\u2460\u2461\u5373\u6570\u5b66\u5f52\u7eb3\u6cd5\u539f\u7406\u5f97\u539f\u547d\u9898\u6210\u7acb\uff0e
(1)a(n+1)=an/(2an+1)
1/a(n+1)=(2an+1)/an=1/an +2
1/a(n+1)-1/an=2，为定值
1/a1=1/1=1，数列{1/an}是以1为首项，2为公差的等差数列
1/an=1+2(n-1)=2n-1
an=1/(2n-1)
n=1时，a1=1/(2×1-1)=1，同样满足通项公式
数列{an}的通项公式为an=1/(2n-1)
(2)
bn=2ⁿ/[1/(2n-1)]=2ⁿ·(2n-1)
Tn=b1+b2+...+bn=1×2+3×2²+5×2³+...+(2n-1)×2ⁿ
2Tn=1×2²+3×2³+...+(2n-3)×2ⁿ+(2n-1)×2^(n+1)
Tn-2Tn=-Tn=2+2²+...+2ⁿ-(2n-1)×2^(n+1)
=2×(2ⁿ-1)/(2-1)-(2n-1)×2^(n+1)
=(1-n)×2^(n+2) -2
Tn=(n-1)×2^(n+2) +2
绛旓細an/n=a(n-1)/(n-1)a1/1=1/1=1 鏁板垪{an/n}鏄悇椤瑰潎涓1鐨勫父鏁版暟鍒椼俛n/n=1 an=n n=1鏃讹紝a1=1锛屽悓鏍锋弧瓒抽氶」鍏紡 鏁板垪{an}鐨勯氶」鍏紡涓篴n=n銆
绛旓細a2/a1=2/1 涓婅堪绛夊紡宸﹁竟涔樹互宸﹁竟锛屽彸杈逛箻浠ュ彸杈规湁 a(n+1)/a1=(n+1)/1 a(n+1)=n+1 鎵浠ャan=n
绛旓細an=a1+(n-1)d =1+(n-1)脳2=2n-1 鎵浠ラ氶」鍏紡涓篴n=2n-1
绛旓細鎵浠2=a1+d=1+d,a5=a1+4d=1+4d,a14=a1+13d=1+13d 鍥犱负{bn}涓虹瓑姣鏁板垪 鎵浠(b3)^2=b2*b4 鍙坅2=b2,a5=b3,a14=b4 鎵浠(a5)^2=a2*a14 鍗(1+4d)^2=(1+d)*(1+13d)鎵浠1+8d+16d^2=1+14d+13d^2 鍗砫^2-2d=0 鎵浠=2鎴杁=0 鍙堝洜涓篸锛0 鎵浠=2 鎵浠an=a1+(...
绛旓細瑙ｇ瓟锛歛2锛a1锛宎3鎴愮瓑宸鏁板垪 鈭 2a1=a2+a3 鍗2a1=a1*q+a1*q²鈭 2=q+q²鈭 q=1鎴杚=-2 鈭 q鈮1 鍒檘=-2 鈭 a1=1,a2=-2,a3=4,a4=-8,a5=16 鈭 S5=1-2+4-8+16=11
绛旓細a1=1 n=1 a2=1/2*1+1/2=1 鍚岀悊 a3=1/2a2+1/2*2=1/2+1/4=3/4
绛旓細S2/S1 = (2+(2-1)d) / a1 = 2+d 鎵浠 n>1 鏃 (2+(2n-1)d)*2 / (2+(n-1)d) = 2+d (2+(2n-1)d)*2 = (2+d) (2+(n-1)d)鏁寸悊寰 (n-1)*d^2 -2nd +2d =0 (n-1)d(d-2) =0 鎵浠 d = 0 鎴 d=2 {an}鐨閫氶」鍏紡涓 an = a1 + (n-1) *0 ...
绛旓細绛夊樊鏁板垪姹傚拰鍏紡 閫氶」鍏紡锛An=A1+(n-1)d An=Am+(n-m)d 绛夊樊鏁板垪鐨鍓峮椤瑰拰锛歋n=[n(A1+An)]/2 Sn=nA1+[n(n-1)d]/2 绛夊樊鏁板垪姹傚拰鍏紡:绛夊樊鏁板垪鐨勫拰=(棣栨暟+灏炬暟)*椤规暟/2;椤规暟鐨勫叕寮:绛夊樊鏁板垪鐨勯」鏁=[(灏炬暟-棣栨暟)/鍏樊]+1....
绛旓細an+1−2n+1锛−(an−2n)锛屾墍浠鏁板垪{an−2n}鏄互a1-2=1涓棣栭」锛屽叕姣斾负-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鎴愮瓑宸暟鍒楋紝鍒欑敱宸茬煡寰楋細2锛2n+锛-1锛塶-1锛=2n-1+锛-1锛塶-...
绛旓細鏁板垪鏄互1涓棣栭」锛3/2涓哄叕姣旂殑绛夋瘮鏁板垪 閫氶」鍏紡涓猴細an=1 (n=1)an=(3/2)^(n-1) (n>1)(2)a2+a4+a6+鈥+a2n 閫氶」鍏紡涓猴細An=(3/2)^(2n-1)棣栭」涓3/2锛屽叕姣斾负(3/2)²鍒檃2+a4+a6+鈥+a2n={(3/2)[1-(3/2)^2n]}/[1-(3/2)²]={(3/2)[1-(...
