高中数列已知{an}的前n项和为Sn Sn=n²+n，{bn}的前n项和为Tn Tn=2bn-2
An = Sn - S(n-1) = 2n;Tn = 2Bn -2 = 2(Tn - T(n-1)) -2 = 2Tn - 2T(n-1) -2;左右两边整理得
Tn+2 = 2(T(n-1) +2);所以Tn+2为等比数列，说以直接代入等比数列公式得Tn=2^(n+1) - 2;也就是2的n+1次方减2了
与An一样，Bn = Tn- T（n-1） =2^n;
An.Bn = n*2^(n+1);
{1/an·an+1} = 1/(2n*2(n+1)) = (1/n-1/(n+1)) / 4；sn一叠加就出来了
4就变成的等比数列，公比为1/4了
5题这样做，将（n+1）项除以第n项，将化简的值与1比较，求出接近1的左右两项，分别算出结果，谁大就是谁了。这样写太不方便了，没办法只能这样表达了
6.7就是将各项分解再错位相减，那样就可以得出了，如果你真的特别想得到结果可以在M我，再详谈
绛旓細瑙ｏ細锛1锛夌敱Sn = 2an鈥2 鍙緱锛屽綋n=1鏃讹紝S1 = a1 = 2 a1鈥2 瑙ｅ緱a1 = 2 鍙圫n -1 = 2an-1鈥2 鍒橲n 鈥 Sn-1 = an = 2an鈥2鈥旓紙2an-1鈥2锛=2an鈥2an-1 鏁寸悊鍙緱锛宎n = 2 an-1 锛屼负绛夋瘮鏁板垪锛屽叕姣斾负q = 2 鏁卆n = a1•qn-1 = 2•2n-1 = 2n...
绛旓細8S1=8a1=a1^2+4a1-5 a1>0 a1=5 n>=2鏃 8sn=an^2+4an-5 8S(n-1)=a(n-1)^2+4a(n-1)-5 an^2-a(n-1)^2-4an-4a(n-1)=0 (an+a(n-1))(an-a(n-1)-4)=0 an>0 an-a(n-1)-4=0 a1=5 an=4n+1 bn=(an-d)an(an+d)=(4n+1)*(16n^2+8n+1-d^2)...
绛旓細(1) -(2)寰 an -a(n-1) +an=2 鍗 2an -a(n-1)=2 2an=a(n-1) +2 2(an -2)=a(n-1) -2 (an - 2)/[a(n-1) -2 ]=1/2 鎵浠 {an -2}鏄叕姣斾负1/2鐨勭瓑姣鏁板垪锛岃岀敱 a1+S1=2锛屽緱 a1=1锛宎1-2=-1 鎵浠 an - 2=-(1/2)^(n-1)锛宎n=2-(1/2)^(...
绛旓細涓妤艰搴熻瘽鐨勯偅浜虹湡璁ㄥ帉锛屾垜灏辩粡甯稿湪缃戜笂闂紝鏈変粈涔堥敊鍟 a(n+1)=3Sn 鈶 an=3S(n-1) 鈶 鑱旂珛涓婁袱寮忚В寰 a(n+1)=4an 濡傛灉an+1=3Sn=1璇存槑an鍜Sn閮芥槸甯告暟鏁板垪 鎵浠ユ棤瑙ｏ紝鎴戞劅瑙夐璨屼技鏈夐敊銆傘傚鏋滅渷鐣=1鐨勬潯浠剁殑璇 杩欏氨鏄竴涓瓑姣旀暟鍒 a(n+1)/an=4 鍙坅1=1 瑙ｅ緱 an=4...
绛旓細(1)Sn=n²+2n 褰搉=1鏃讹紝a1=S1=3 褰搉鈮2鏃讹紝an=Sn-S(n-1)=n²+2n-(n-1)²-2(n-1)=2n+1 涓婂紡瀵筺=1涔熸垚绔 鍥犳閫氶」鍏紡涓 an=2n+1 (2)Tn=1/a1a2+1/a2a3+1/a3a4+L+1/anan+1 =1/(3*5)+1/(5*7)+1/(7*9)1+...+1/[(2n+1)(2n++3)]...
绛旓細n-1)锛屽洜涓篠n-Sn-1=An锛屽悗闈㈢殑灏卞彲浠ョ畻鍑烘潵浜嗭紝绗簩闂繕娌＄畻绗簩灏忛棶:鍥犱负鏁板垪bn鏄瓑姣旀暟鍒,鎵浠2/b1=q,锛屽張鍥犱负b1=a1,b2=a2銆傛墍浠=a2/a1锛屽洜涓篴k=b11锛屾墍浠k=b1Xq^10=a1X(a2/a1)^10锛岀劧鍚庝綘鑷繁绠楀嚭a1锛宎2鐨勫煎氨鍙互寰楀嚭ak鐨勫硷紝鐒跺悗灏哸k鐨勫间唬鍏an=2n锛屽氨鍙互绠楀嚭k浜嗭紒
绛旓細Sn+1+(5n+2)Sn=-20 (5n-3)an+2-(5n+2)an+1=20 鍒 (5n+2)an+3-(5n+7)an+2=20 涓ゅ紡鐩稿噺,寰楋細(5n+2)an+3-(10n+4)an+2+(5n+2)an+1=0 an+3-2an+2+an+1=0 鍙宸茬煡a1=1,a2=6,a3=11锛岀患涓婏紝an+2-2an+1+an=0鍗2an+1=an+an+2 璇佸緱{an}涓虹瓑宸鏁板垪 ...
绛旓細瑙ｏ細褰搉=1鏃讹紝a1=s1=2/1=2 褰搉>=2鏃讹紝an=sn-s(n-1)=(n+1)/n-n/(n-1)=1/[n(1-n)]
绛旓細2Sn=(n+2)an-1 锛屾墍浠2S(n-1)=(n+1)a(n-1)-1,涓ゅ紡鐩稿噺寰2an=(n+2)an-(n+1)a(n-1),鎵浠n/a(n-1)=(n+1)/n,鎵浠n=an/a(n-1)*a(n-1)/a(n-2)*:::*a2/a1*a1=(n+1)/2,鎵浠1/ana(n+2)=4/((n+1)(n+3))=2*(1/(n+1)-1/(n+3)),Tn=2(1/...
绛旓細an=S(n+1)-Sn=n+1 鑻涓哄伓鏁,Tn=(b1+b3+...+b(n-1))+(b2+b4+..+bn)=n(2+n)/4+(2^(n+2)-4)/(1-1/4)=n(n+2)/4+(2^(n+4)-16)/3=n(n+2)/4+16(2^n-1)/3.鑻涓哄鏁癟n=(b1+b3+...+bn)+(b2+b4+..+b(n+1))-b(n+1)=(n+1)(n+3)/4+16...
