已知数列{an},an+1=an²,a1=1,求an的通项 已知数列{an}中，a1=1,a(n+1)=an/(2an+...
\u5df2\u77e5\u6570\u5217{an}\u6ee1\u8db3an+1=an+2n+1\uff0ca1=1\uff0c\u6c42\u6570\u5217{an}\u7684\u901a\u9879\u516c\u5f0f\u7531an+1=an+2n+1\u5f97an+1-an=2n+1\u5219an=\uff08an-an-1\uff09+\uff08an-1-an-2\uff09+\u2026+\uff08a3-a2\uff09+a1=[2\uff08n-1\uff09+1]+[2\uff08n-2\uff09+1]+\u2026+\uff082\u00d72+1\uff09+\uff082\u00d71+1\uff09+1=2[\uff08n-1\uff09+\uff08n-2\uff09+\u2026+2+1]+\uff08n-1\uff09+1=2\u00d7(n?1)n2+\uff08n-1\uff09+1=\uff08n-1\uff09\uff08n+1\uff09+1=n2\uff0c\u6240\u4ee5\u6570\u5217{an}\u7684\u901a\u9879\u516c\u5f0f\u4e3aan=n2\uff0e
a(n+1)=an/(2an +1)
1/a(n+1)=(2an +1)/an =1/an +2
1/a(n+1)-1/an=2\uff0c\u4e3a\u5b9a\u503c\u3002
1/a1=1/1=1
\u6570\u5217{1/an}\u662f\u4ee51\u4e3a\u9996\u9879\uff0c2\u4e3a\u516c\u5dee\u7684\u7b49\u5dee\u6570\u5217\u3002
1/an =1+2(n-1)=2n-1
an=1/(2n-1)
\u6570\u5217{an}\u7684\u901a\u9879\u516c\u5f0f\u4e3aan=1/(2n-1)\u3002
a1=1
a2=1²=1
a3=a2² =1
...............
an=1
绛旓細鎵浠an閫氶」鍏紡涓篈脳1^n+B脳(-1/2)^n A,B涓哄緟瀹氱郴鏁 a1=A-B/2=1 a2=A+B/4=2 寰 A=5/3 B=4/3 an=[5+4脳(-1/2)^n]/3 鑻ユ病鏈夊杩囩壒寰佹柟绋嬶紝鍙涓嬭浆鎹 a[n+2]-a[n+1]=-(a[n+1]-a[n])/2 绛夋瘮鏁板垪 鎵浠[n+2]-a[n+1]=锛-1/2锛塣n (...
绛旓細鍥燽n=a(n+1)-2an=4[an-a(n-1)]-2an=2an-4a(n-1)=2*b(n-1)鎵浠ワ細bn鏄叕姣斾负2鐨勭瓑姣鏁板垪锛鐢盿1=1,s2=4a1+2,鐭2=5,浠庤宐1=a2-2a1=5-2脳1=3 鍥犳bn=3*2^(n-1)2锛夎cn=an/2^n锛屾眰璇乧n鏄瓑宸暟鍒 鐢眂n=an/2^n锛岀煡an=2^n*cn,涓攁(n+1)=2^(n+1)*c(n+1...
绛旓細涓よ竟闄や互2^n an/2^n=2a(n-1)/2^n+1 an/2^n=a(n-1)/2^(n-1)+1 an/2^n-a(n-1)/2^(n-1)=1 鎵浠n/2^n鏄瓑宸紝d=1 鎵浠n/2^n=a1/2^1+d(n-1)an/2^n=1/2+n-1 an=(n-1/2)*2^n
绛旓細瑙ｏ細鐢遍鍙緱锛歛1=1 a2=a1+2x1+5 a3=a2+2x2+5 鈥︹an=a(n-1)+2x(n-1)+5 鎶婁互涓婃墍鏈夊紡瀛愮浉鍔犲苟鍖栫畝寰楋細an=2x[1+2+3+鈥︹+(n-1)]+5x(n-1)+1=n(n-1)+5n-4=n�0�5+4n-4
绛旓細鍥犱负a(n+2)+a(n)=2路a(n+1)鎵浠(n+2)-a(n+1)=a(n+1)-a(n)鎵浠鏁板垪{an}鏄瓑宸暟鍒 鍥犱负a1=1锛a2=3 鎵浠ョ瓑宸暟鍒梴an}鐨勫叕宸甦=2 鎵浠ユ暟鍒梴an}鐨勫墠n椤瑰拰鍏紡Sn=(1/2)路[1+(2n-1)]路n=n²
绛旓細[a(n+i)+3^n]/(a(n)+3^n]=2 鍙緱鍑猴經a(n)+3^n锝濇槸棣栭」涓4锛屽叕姣斾负2鐨勭瓑姣鏁板垪銆戞槸閿欑殑 姝ｈВ锛氣埖a(n+1)=2an+3^n,鈭碼(n+1)- 3^(n+1)=2(an-3^n)鈭碵a(n+1)-3^(n+1)]/(an-3^n)=2 鈭锝沘(n)-3^n锝濇槸棣栭」涓-2锛屽叕姣斾负2鐨勭瓑姣旀暟鍒 鈭碼n-3^n=-2*2...
绛旓細a1a2a3鈥︹n=n^2 a1a2a3鈥︹anan+1=(n+1)^2 涓嬪紡/涓婂紡=an+1=(n+1)²/n²an=n²/(n-1)² (a鈮2)a1=1,1,
绛旓細瑙ｏ細a(n+1)=3an+2^(n+1)璁綼(n+1)+k*2^(n+1)=3[a(n)+k*2^n]鍒 a(n+1)=3an+3k*2^n-2k*2^n=3an+k*2^n 鎵浠 k=2 鍗 a(n+1)+2*2^(n+1)=3[a(n)+2*2^n]鎵浠 {an+2*2^n}鏄瓑姣鏁板垪 棣栭」涓篴1+2*2^1=5,鍏瘮涓3 鎵浠 an+2*2^n=5*3^(n-...
绛旓細渚濋鎰忓彲鐭ワ細an-a(n-1)=2锛屾墍浠鏁板垪an鏄叕宸甦=2鐨勭瓑宸暟鍒 渚濇嵁绛夊樊鏁板垪鐨勯氶」鍏紡鍙煡 an=a1+(n-1)d =1+(n-1)脳2=2n-1 鎵浠ラ氶」鍏紡涓篴n=2n-1
绛旓細鍏瘮涓4鐨勭瓑姣鏁板垪.a(n+1)-an=4*4^(n-1)=4^n 鈥︹ 鈥︹2-a1=4^1 宸﹀彸绱姞 a(n+ 1)-a1=[4^(n)+4^(n-2)+鈥︹+4^1]=4(4^n-1)/(4-1)=4/3*(4^n-1)a(n+1)=4/3*(4^n-1)+a1=4/3*4^n-1/3 an鐨勮〃杈惧紡 a(n)=4/3*4^(n-1)-1/3 ...
