急急急！高一数学超简单数列问题！学霸秒杀不用思考！
\u9ad8\u4e00\u6570\u5b66\u7b80\u5355\u6570\u5217\u9898\u2235S(2n-1)=(2n-1)(A1+A(2n-1))/2=(2n-1)(2An)/2=An(2n-1)
\u540c\u7406 \uff1aT\uff082n-1\uff09\uff1dBn\uff082n-1\uff09
\u2234S(2n-1)/T\uff082n-1\uff09\uff1dAn/Bn\uff1d2\uff082n-1\uff09/{3(2n-1)+1}=(2n-1)/(3n-1)
\u82e5\u6709\u4e0d\u61c2\u53ef\u518d\u95ee\u6211\u3002
1/(2X5)+1/(5X8)+1/(8X11)+...+1/((3n-1)*(3n+2))
=1/3*(1/2-1/5+1/5-1/8+1/8-1/11+...+1/(3n-1)-1/(3n+2))
=1/3*(1/2-1/(3n+2))
=n/(6n+4)
2^2/(1X3)+4^2/(3X5)+6^2/(5X7)+...+(2n)^2/((2n-1)*(2n+1))
\u56e0\u4e3a
(2n)^2/((2n-1)*(2n+1)
=((2n)^2-1+1)/((2n)^2-1)
=1+1/((2n)^2-1)
=1+1/2*(1/(2n-1)-1/(2n+1))
\u6240\u4ee5\u539f\u5f0f=
1+1/2*(1/1-1/3)+1+1/2*(1/3-1/5)+1+1/2*(1/5-1/7)+...+1+1/2*(1/(2n-1)-1/(2n+1))
=1*n+1/2*(1-1/3+1/3-1/5+1/5-1/7+...+1/(2n-1)-1/(2n+1))
=n+1/2*(1-1/(2n+1))
=(2n^2+2n)/(2n+1)
（n+3）a（n+1）²-nan²+3a（n+1）an=0
=[（n+3）a（n+1）-nan]*[a（n+1）+an] =0
所以：（n+3）a（n+1）-nan = 0
a（n+1）/ an = n/ (n+3)
an/ a1 = 6 /n（n+1）（n+2）
an=2 /n（n+1）（n+2）
绛旓細绛旀锛1銆2n-1 2銆乶/[2n+1]璇峰弬鐪嬩笅鍥撅細
绛旓細瑙ｆ瀽锛氳鏁板垪{an}鐨勫叕宸负d锛岄偅涔堬細a3=a1+2d锛宎4=a1+3d锛宎5=a1+4d 鑻1锛宎3锛宎4锛宎5缁勬垚涓涓瓑姣旀暟鍒楋紝閭ｄ箞锛歛3/a1=a5/a4鍗砤3*a4=a1*a5 鎵浠ワ細(a1+2d)(a1+3d)=a1*(a1+4d)a1²+5a1*d+6d²=a1²+4a1*d 6d²+a1*d=0 d(6d+a1)=0 鐢变簬d鈮0锛屾墍浠ヨВ寰...
绛旓細a1,a3,a9鎴愮瓑姣鏁板垪 (A3)^2=A1*A9=A9 A3=A1+2d=1+2d A9=A1+8d=1+8d (1+2d)^2=1+8d d=1 An=A1+(n-1)d=n Sn=2*(A1+A2+...+An)=2*(A1+An)*n/2 =n(n+1)甯屾湜鎴戠殑鍥炵瓟鑳藉府鍔╁埌鎮紝婊℃剰鐨勮瘽鐑﹁閲囩撼~
绛旓細(1)a1=3/5 a(n+1)= 3an/(4an+1)1/(an+1) = 4/3 + (1/3)(1/an)1/(a(n+1) - 2 = (1/3) ( 1/an - 2)=> {1/an - 2} 鏄瓑姣鏁板垪 , q= 1/3 (2)1/an - 2 = (1/3)^(n-1) .( 1/a1 - 2 )=2 .(1/3)^n 1/an = 2 + 2 .(1/3)^n...
绛旓細a1=1, a(n+1)=an+位.2^n a1,a2+2, a3 is a AP (1)a1+a3 = 2(a2+2)1+ a2+4位 =2(a2+2)1+ 4位 = a2+4 = 1+2位 + 4 位 =2 (2)a(n+1)=an+2^(n+1)a(n+1) -an = 2^(n+1)an -a1 = 2^2+2^3+...+2^n = 4(2^(n-1) -1)an = 2^(n...
绛旓細s2=2a2+1 s2=a1+a2 鎵浠2=0 s3=2a3-1 s3=a1+a2+a3=1+a3 鎵浠3=2 (2)n涓哄伓 sn=2an+1 s(n-1)=2a(n-1)+1 涓ゅ紡鐩稿噺 寰 2a(n-1)=an锛堜负绛夋瘮鏁板垪锛夐椤逛负a2=0 鎵浠ュ綋n涓哄伓鏃 an=0 n涓哄 sn=2an-1 s(n-1)=2a(n-1)-1 鍚屾牱寰楀埌 2a(n-1)=an锛堜负绛夋瘮...
绛旓細an=1+鈭2+2(n-1)=2n+鈭2-1 Sn=n(a1+an)锛2=n²+鈭2n bn=n+鈭2 璁撅箼i+鈭2锕²=锕檍+鈭2锕氾箼k+鈭2锕歩²+2鈭2i+2=jk+鈭2锕檍+k)+2 i²-jk=鈭2锕檍+k-2i)鈭2=锕檌²-jk)锛忥箼j+k-2i)鐭涚浘 鈭达箾bn锕滅殑浠绘剰3椤逛笉鍙兘鎴愪负绛夋瘮鏁板垪 ...
绛旓細绗竴棰橈細鍏堝垎鏋愭瘡涓骞村瓨娆剧殑鏈埄鍜岋紝灏忚姵鍚屽涓骞磋瀛樻12娆★紝姣忔瀛樻5鍏冿紝鍚勬瀛樻鍙婂叾鍒╂伅鎯呭喌濡備笅锛氱12娆″瓨娆5鍏冿紝杩欐椂瑕佸埌鏈熸敼瀛橈紝鍥犳杩欐鐨勫瓨娆炬病鏈夋湀鎭紱绗11娆″瓨娆5鍏冿紝杩1涓湀鍗冲埌鏈燂紝鍥犳鎵瀛樻涓庡埄鎭箣鍜屼负锛5锛5脳0.2%=5脳(1锛0.2%)锛涚10娆″瓨娆5鍏冿紝杩2涓湀鍒版湡锛屽洜姝ゅ瓨娆句笌...
绛旓細鐢盿2+a4=b3,b2b4=a3寰楋細{2+4d=2*q^2 {4*q^4=1+2d 瑙ｅ緱锛歞=-3/8,q=1/2 鏁咃紝S10=a1*n+n(n-1)d/2=-55/8 T10=a1(1-q^n)/(1-q) =4-(1/2)^(n-2)鐢变簬鏃堕棿鍘熷洜灏辩粰鎮ㄨВ绛旂涓棰樺惂锛岃В棰樻濊矾灏辨槸杩欐牱鐨勶紝鍙兘鍦ㄨВ棰樿繃绋嬩腑鏈変簺璁＄畻閿欒锛屾偍鑷繁鍙互閲嶆柊鍐嶇畻涓閬嶃
绛旓細bn=1/3*b(n-1)锛屾晠鏁板垪{bn}鏄瓑姣旀暟鍒椼備笖鐢眀1+2S1-2=0锛屽彲寰梑1+2b1-2=0锛宐1=2/3锛屾暟鍒梴bn}鐨勯氶」鍏紡鏄 bn=b1*(1/3)^(n-1)=2/3^n銆3) 鐢遍鎰忥紝cn=an*bn=(n+1)*2/3^n, 鎵浠ユ湁 Tn = 2*2/3^1+2*3/3^2+...+2*n/3^(n-1)+2*(n+1)/3^n锛屽洜姝 1/...
