数列an前n项和为Sn，且sn十an=1，求an通项公式
\u5df2\u77e5\u6570\u5217{an}\u7684\u524dN\u9879\u548c\u4e3aSn \u4e14an+1=Sn-n+3,a1=2,.\u6c42an\u7684\u901a\u9879\u516c\u5f0f\u4f60\u597d
\u6211\u8ba4\u4e3a\u95ee\u9898\u51fa\u5728n\u7684\u53d6\u503c\u8303\u56f4\u4e0a
an+1=Sn-n+3,\u6b64\u65f6n\u7684\u53d6\u503c\u8303\u56f4\u662fn\u22651
an=Sn-1-n+4\u6b64\u65f6n\u7684\u53d6\u503c\u8303\u56f4\u662fn\u22652
\u6240\u4ee5\u4e24\u8005\u5fc5\u987b\u5148\u7edf\u4e00n\u7684\u8303\u56f4\u624d\u80fd\u591f\u76f8\u51cf
\u6240\u4ee5\u5e94\u8be5\u4e3a2an-1=an+1(n\u22652\uff09
\u6545\u4e0d\u80fd\u5c06n=1\u4ee3\u5165\u4e0a\u5f0f\u3002
2an-1=an+1(n\u22652\uff09
\u5373(an+1-1)-2(an-1)=0
\u4ee4an-1=bn
\u5219bn+1-2bn=0
(bn+1)/bn=2
\u6545bn\u662f\u7b49\u6bd4\u6570\u5217
b1=a1-1=1
\u6240\u4ee5bn=2^(n-1)
\u6240\u4ee5an=bn+1=2^(n-1)+1(n\u22652\uff09
\u8fd9\u65f6\u624d\u80fd\u5c06n=1\u4ee3\u5165\u68c0\u9a8c
a1=1+1=2\uff0c\u7b26\u5408\u4e0a\u5f0f
\u6240\u4ee5an=2^(n-1)+1
1\u3001S(n+1)-Sn=(1/3)^n+1=a(n+1)
an=(1/3)^(n-1)+1 n\u22652 a1=1/3
Sn=1/3+1/3+1+(1/3)^2+1+.......+(1/3)^(n-1)+1
=1/3+(n-1)+[1-(1/3)^(n-1)]/2
=n-2/3+[1-(1/3)^(n-1)]/2
2\u3001S1=1/3
t(S1+S2)=2t
3(S2+S3)=40/3
S1\uff0ct(S1+S2),3(S2+S3)\u6210\u7b49\u5dee\u6570\u5217\uff0c
\u6240\u4ee5 2*2t=1/3+40/3
t=41/12
∴S(n-1)+a(n-1)=1
两式相减得：
an+an-a(n-1)=0
即an/a(n-1)=1/2
故an是以a1为首项,1/2为公比的等比数列：
且S1+a1=1 即a1=1/2
故an=a1*q^(n-1)
=1/2^n
如有不懂，可追问！
解：当n=1时，a1=0.5
当n=2时，带入a1得a2=1/4
同理当n=3时,n=1/8
可以看出an=2^(-n)
当n=k时，sk+ak=1,当n=k+1时
s(k+1)+a(k+1)=sk+a(k+1)+a(k+1)=sk+2a(k+1)=sk+2*2^(-k-1)=sk+2^(-k)=sk+ak=1成立
故an=2^(-n)
绛旓細n=1 , a1= 2+a an = Sn-S(n-1)= 2^(n-1)a1= 1= 2+a a=-1 (2)bn=nan =n.2^(n-1)let S=1.2^0+2.2^1+...+n.2^(n-1) (1)2S= 1.2^1+2.2^2+...+n.2^n (2)(2)-(1)S = n.2^n -[1+2+...+2^(n-1)]=n.2^n - (2^n -1)...
绛旓細=2an--2an--1 an=2an--1 鎵浠 an/an--1=2,鍗 鍏瘮q=2,鎵浠 鏁板垪{an}鏄瓑姣鏁板垪锛鍥犱负 a1=S1=2a1--4,鎵浠 a1=4,鎵浠 閫椤筧n=a1q^(n--1)=4*2^(n--1)=2^(n+1)鍗 an=2^(n+1).鎵浠 Sn=a1(q^n--1)/(q--1)=4(2^n--1)/(2--1)...
绛旓細鈭碼1=S1=3a1+1锛宎1=-1/3 涓擲(n-1)=3a(n-1)+1 鈭an=Sn-S(n-1)=3an-3a(n-1)鈭2an=3a(n-1)鈭碼n=(3/2)a(n-1)鈭碼n鏄互a1=-1/3涓洪椤癸紝q=3/2涓哄叕姣旂殑绛夋瘮鏁板垪 鈭碼n=-1/3脳(3/2)^(n-1)寰堥珮鍏翠负鎮ㄨВ绛旓紝绁濅綘瀛︿範杩涙锛併愬涔犲疂鍏搞戝洟闃熶负鎮ㄧ瓟棰樸傛湁涓嶆槑鐧界殑鍙互...
绛旓細n=1鏃讹紝a1=S1=4a1-a2 a2=3a1=3 n鈮2鏃锛孲n=4an-a(n+1)S(n+1)=4a(n+1)-a(n+2)a(n+1)=S(n+1)-Sn=4a(n+1)-a(n+2)-4an+a(n+1)[a(n+2)-2a(n+1)]/[a(n+1)-2an]=2锛屼负瀹氬 a2-2a1=3-2=1锛屾暟鍒{a(n+1)-2an}鏄互1涓洪椤癸紝2涓哄叕姣旂殑绛夋瘮鏁板垪 a...
绛旓細锛2)鐢an=Sn-S(n-1)鍙互寰楋紙an+a(n-1))(an-a(n-1)-2)=0,鍙堝洜涓篴n銆0锛屾墍浠n=a(n-1)+2,鎵浠(n-1)=a(n-2)+2...a2=a1+2锛屽皢涓婇潰鎵鏈夌瓑寮忕浉鍔犲彲浠ュ緱an=1+2锛坣-1锛夛紙3锛塨n=21-2n,浠n>0,鍙互寰梟<10.5,鎵浠ュ綋n=10鏃禸n鐨鍓峮椤瑰拰鏈澶 甯屾湜鑳藉府鍒颁綘 ...
绛旓細瑙ｏ細鈭电瓑姣旀暟鍒梴an}鐨鍓峮椤瑰拰涓篠n锛骞朵笖瀵逛换鎰忔鏁存暟n閮芥湁Sn+2=4Sn+3鎴愮珛锛屸埓Sn+1=4Sn-1+3涓嶴n+2=4Sn+3涓ゅ紡鐩稿噺寰楀嚭锛歛n+2=4an 鍗砤n+2an=4锛屸埖绛夋瘮鏁板垪{an} 鈭磓2=4锛宷=卤2锛屽綋q=2鏃讹紝a1脳1-2n+21-2=4a1脳1-2n1-2+3锛宎1=1锛屸埓a2=2锛屽綋q=-2鏃讹紝a1脳1-(-2)n+...
绛旓細锛1锛Sn=2^n+a 褰搉=1鏃讹紝A1=2+a 褰搉=2鏃讹紝A1+A2=S2=4+a 閭ｄ箞A2=2 褰搉=3鏃讹紝A1+A2+A3=S3=8+a 閭ｄ箞A3=4 鈭祘An}涓虹瓑姣旀暟鍒 鈭碅2/A1=A3/A2 鈭2/(2+a)=4/2=2 鈭碼=-1 A1=1,鍏瘮q=2 An=2^(n-1)(2)Bn=(2n-1)2^(n-1)Tn=1+3*2+5*2^2+...+(2n-1)2...
绛旓細璇佹槑 鐢 sn+n=2an 鍙緱 S(n+1)+n+1=2a(n+1) a1=1 涓ゅ紡鐩稿噺寰 S(n+1)-Sn +1=2a(n+1)-2an 鍗砤(n+1)+1=2a(n+1)-2an a(n+1)=2an+1 a(n+1锛+1=2(an+1)a(n+1)+1/(an+1)=2 鏁板垪锕沘n+1锕滄槸鍏瘮涓2鐨勭瓑姣旀暟鍒 ...
绛旓細鍥犱负Sn=a1+a2+鈥︹+an-4+an-3+an-2+an-1+an Sn-4=a1+a2+鈥︹+an-4 鎵浠n-Sn-4=(a1+a2+鈥︹+an-4+an-3+an-2+an-1+an)-(a1+a2+鈥︹+an-4)=an+an-1+an-2+an-3 鍙圫n=210锛孲n-4=130 鎵浠n-Sn-4=210-130=80 鎵浠n-Sn-4=an+an-1+an-2+an-3=80 ...
绛旓細an=Sn-S(n-1)=n²+4n-(n-1)²-4(n-1)=2n+3 n=1鏃讹紝a1=2+3=5锛屽悓鏍锋弧瓒炽傛暟鍒梴an}鐨勯氶」鍏紡涓篴n=2n+3銆(2)bn=(an+5)脳2^(n-1)=(2n+3+5)脳2^(n-1)=(2n+8)脳2^(n-1)=(n+4)脳2ⁿ=n脳2ⁿ+2^(n+2)Tn=b1+b2+...+bn =(1脳2...
