å
¬å¼æ³ãç´¯å æ³ãç´¯ä¹æ³ãå¾
å®ç³»æ°æ³ã对æ°åæ¢æ³ãè¿ä»£æ³ãæ°å¦å½çº³æ³ãæ¢å
æ³ãä¸å¨ç¹æ³ãç¹å¾æ ¹çæ¹æ³ççã
ç±»åä¸
å½çº³âçæ³âè¯æ
ç±æ°åçéæ¨å
¬å¼å¯ååºæ°åçåå 项ï¼åç±åå 项æ»ç»åºè§å¾ï¼çæ³åºæ°åçä¸ä¸ªé项å
¬å¼ï¼æåç¨æ°å¦å½çº³æ³è¯æï¼
ç±»åäº
âéå·®æ³âåâ积åæ³â
(1)å½æ°åçéæ¨å
¬å¼å¯ä»¥å为an+1-an=f(n)æ¶ï¼ån=1,2,3,â¦,n-1ï¼å¾n-1个å¼åï¼
a2-a1=f(1),a3-a2=f(2),â¦,an-an-1=f(n-1)ï¼
ä¸f(1)+f(2)+â¦+f(n-1)å¯æ±å¾æ¶ï¼ä¸¤è¾¹ç´¯å å¾é项anï¼æ¤æ³ç§°ä¸ºâéå·®æ³âï¼
(2)å½æ°åçéæ¨å
¬å¼å¯ä»¥å为an+1/an=f(n)æ¶ï¼ä»¤n=1,2,3,â¦,n-1,å¾n-1个å¼åï¼å³
a2/a1=f(1),a3/a2=f(2),a4/a3=f(3),â¦,an/an-1=f(nï¼1)ï¼ä¸f(1)f(2)f(3)â¦f(n-1)å¯æ±å¾æ¶ï¼ä¸¤è¾¹è¿ä¹å¯æ±åºanï¼æ¤æ³ç§°ä¸ºâ积åæ³âï¼
ç±»åä¸
æé æ³
éæ¨å¼æ¯pan=qan-1+f(n)(pãqæ¯ä¸ä¸ºé¶ç常æ°)ï¼å¯ç¨å¾
å®ç³»æ°æ³æé ä¸ä¸ªæ°ççæ¯æ°åæ±è§£ï¼
ç±»åå
å¯è½¬å为类åä¸æ±é项
(1)â对æ°æ³â转å为类åä¸ï¼
éæ¨å¼ä¸ºan+1=qan�k(qï¼0,kâ 0ä¸kâ 1,a1ï¼0)ï¼ä¸¤è¾¹å常ç¨å¯¹æ°ï¼å¾lgan+1=klgan+lgqï¼ä»¤lgan=bnï¼åæbn+1=kbn+lgqï¼è½¬å为类åä¸ï¼
(2)âåæ°æ³â转å为类åä¸ï¼
éæ¨å¼ä¸ºåçå½¢å¼ï¼an+1=(pan+b)/(qan+c)(anâ 0ï¼pqâ 0,pcâ qb)ï¼
è¥b=0ï¼å¾an+1=pan/(qan+c)ï¼å 为anâ 0ï¼æ以两边ååæ°å¾1/an+1=q/p+c/pan,令bn=1/an,åbn+1=(c/p)bn+q/pï¼è½¬å为类åä¸ï¼
è¥bâ 0ï¼è®¾an+1+x=y(an+x)/qan+cï¼ä¸å·²ç¥éæ¨å¼æ¯è¾æ±å¾xãyï¼ä»¤bn=an+xï¼å¾bn+1=ybn/qan+cï¼è½¬å为b=0çæ
åµï¼
ç±»åäº
éæ¨å¼ä¸ºan+1/an=qn/n+k(qâ 0,kâN)
å¯å
å°çå¼(n+k)an+1=qnan两边åä¹ä»¥(n+k-1)(n+k-2)â¦(n+1)ï¼å¾(n+k)(n+k-1)(n+k-2)⦠(n+1)an+1=q(n+k-1)(n+k-2)â¦(n+1)nanï¼ä»¤bn=(n+k-1)(n+k-2)â¦(n+1)�6�1nan,åbn+1= (n+k)(n+k-1)(n+k-2)â¦(n+1)an+1ï¼
ä»èbn+1=qbnï¼å æ¤æ°åï½bnï½æ¯å
¬æ¯ä¸ºqï¼é¦é¡¹ä¸ºb1=k(k-1)(k-2)â¦2�6�11�6�1a1=k!a1ççæ¯æ°åï¼è¿èå¯æ±å¾anï¼
æ»ä¹ï¼ç±æ°åçéæ¨å
¬å¼æ±é项å
¬å¼çé®é¢æ¯è¾å¤æï¼ä¸å¯è½ä¸ä¸è®ºåï¼ä½åªè¦æ们æä½éæ¨æ°åçéæ¨å
³ç³»ï¼åæç»æç¹å¾ï¼åäºåçåå½¢ï¼å°±è½æ¾å°è§£å³é®é¢çææéå¾ï¼
ç±»åä¸�å½çº³âçæ³âè¯æ
ç±æ°åçéæ¨å
¬å¼å¯ååºæ°åçåå 项ï¼åç±åå 项æ»ç»åºè§å¾ï¼çæ³åºæ°åçä¸ä¸ªé项å
¬å¼ï¼æåç¨æ°å¦å½çº³æ³è¯æï¼
�ä¾1�设æ°åï½anï½æ¯é¦é¡¹ä¸º1çæ£é¡¹æ°åï¼ä¸(n+1)a2n+1-nan2+an+1an=0(n=1,2,3,â¦)ï¼åå®çé项å
¬å¼æ¯an=______________ï¼(2000å¹´å
¨å½æ°å¦å·ç¬¬15é¢)
解ï¼å°(n+1)a2n+1-nan2+an+1an=0(n=1,2,3,â¦)å解å å¼å¾(an+1+an)ã(n+1)an+1-nanã=0ï¼
��ç±äºanï¼0ï¼æ
(n+1)an+1=nanï¼å³an+1=n/(n+1)anï¼
��å æ¤a2=(1/2)a1=(1/2)ï¼a3=(2/3)a2=(1/3),â¦ï¼çæ³an=(1/n),å¯ç±æ°å¦å½çº³æ³è¯æä¹ï¼è¯æè¿ç¨ç¥ï¼
ç±»åäº�âéå·®æ³âåâ积åæ³â
(1)å½æ°åçéæ¨å
¬å¼å¯ä»¥å为an+1-an=f(n)æ¶ï¼ån=1,2,3,â¦,n-1ï¼å¾n-1个å¼åï¼
a2-a1=f(1),a3-a2=f(2),â¦,an-an-1=f(n-1)ï¼
ä¸f(1)+f(2)+â¦+f(n-1)å¯æ±å¾æ¶ï¼ä¸¤è¾¹ç´¯å å¾é项anï¼æ¤æ³ç§°ä¸ºâéå·®æ³âï¼
ä¾2�å·²ç¥æ°åï½anï½æ»¡è¶³a1=1,an=3n-1+an-1(nâ¥2),è¯æï¼an=(3n-1)/2ï¼
(2003å¹´å
¨å½æ°å¦å·æç§ç¬¬19é¢)
è¯æï¼ç±å·²ç¥å¾an-an-1=3n-1ï¼æ
an=(an-an-1)+(an-1ï¼an-2)+â¦+(a2-a1)+a1=3n-1+3��n-2�+â¦+3+1=3n-1/2ï¼
æ以å¾è¯ï¼
(2)å½æ°åçéæ¨å
¬å¼å¯ä»¥å为an+1/an=f(n)æ¶ï¼ä»¤n=1,2,3,â¦,n-1,å¾n-1个å¼åï¼å³
a2/a1=f(1),a3/a2=f(2),a4/a3=f(3),â¦,a��n�/an-1�=f(nï¼1)�ï¼�ä¸f(1)f(2)f(3)â¦f(n-1)å¯æ±å¾æ¶ï¼ä¸¤è¾¹è¿ä¹å¯æ±åºanï¼æ¤æ³ç§°ä¸ºâ积åæ³âï¼
ä¾3�(åä¾1)(2000å¹´å
¨å½æ°å¦å·ç¬¬15é¢)
å¦è§£ï¼å°(n+1)a2n+1-nan2ï¼an+1an=0(n�=1,2,3,â¦)åç®ï¼å¾(n+1)an+1=nanï¼å³
an+1/an=n/(n+1)�
æ
an=an/an-1�6�1an-1/an-2�6�1an-2/an-3�6�1â¦�6�1a2/a1�=n-1/n�6�1n-2/n-1�6�1n-3/n-2�6�1 ⦠�6�11/2�=1/nï¼
ç±»åä¸�æé æ³
éæ¨å¼æ¯pan=qan-1+f(n)(pãqæ¯ä¸ä¸ºé¶ç常æ°)ï¼å¯ç¨å¾
å®ç³»æ°æ³æé ä¸ä¸ªæ°ççæ¯æ°åæ±è§£ï¼
ä¾4�(åä¾2)(2003å¹´å
¨å½æ°å¦å·æç§ç¬¬19é¢)
å¦è§£ï¼ç±an=3n-1+an-1å¾3�6�1an/3n=an-1/3n-1+1ï¼
令bnï¼an/3nï¼åæ
bn=1/3bn-1+1/3ï¼ (*)
设bn+x=1/3(bn-1+x)ï¼åbn=1/3bn-1+1/3x-xï¼ä¸(*)å¼æ¯è¾ï¼å¾x=-1/2ï¼æ以bn-1/2=1/3(bn-1-1 /2)ï¼å æ¤æ°åï½bn-1/2ï½æ¯é¦é¡¹ä¸ºb1-1=a1/3=-1/6ï¼å
¬æ¯ä¸º1/3ççæ¯æ°åï¼æ以bn-1/2=-1/6�6�1(1/3)n-1ï¼å³ an/3n-1/2=-1/6(1/3)n-1ï¼æ
an=3nã1/2-1/6(1/3)n-1ã=3n-1/2ï¼
ä¾5�æ°åï½anï½ä¸ï¼a1=1,an+1=4an+3n+1ï¼æ±anï¼�
解ï¼ä»¤an+1+(n+1)x+y=4(an+nx+y)ï¼å
an+1=4an+3nx+3y-x,ä¸å·²ç¥an+1=4an+3n+1æ¯è¾ï¼å¾
3x=3ï¼ æ以
x=1,
3y-x=1ï¼ y=(2/3)ï¼
æ
æ°åï½an+n+(2/3)ï½æ¯é¦é¡¹ä¸ºa1+1+(2/3)=(8/3)ï¼å
¬æ¯ä¸º4ççæ¯æ°åï¼å æ¤an+n+(2/3)=(8/3)�6�14n-1,å³
an=(8/3)�6�14n-1-n-(2/3)ï¼
å¦è§£ï¼ç±å·²ç¥å¯å¾å½nâ¥2æ¶ï¼an=4an-1+3(n-1)+1ï¼ä¸å·²ç¥å
³ç³»å¼ä½å·®ï¼æan+1-an=4(an-an-1)+3ï¼å³an+1- an+1=4(an-an-1+1)ï¼å æ¤æ°åï½an+1-an+1ï½æ¯é¦é¡¹ä¸ºa2-a1+1=8-1+1=8ï¼å
¬æ¯ä¸º4ççæ¯æ°åï¼ç¶åå¯ç¨âéå·®æ³â æ±å¾å
¶é项an=(8/3)�6�14n-1-n-(2/3)ï¼
ç±»åå�å¯è½¬å为
ç±»åä¸æ±é项
(1)â对æ°æ³â转å为
ç±»åä¸ï¼
éæ¨å¼ä¸ºan+1=qan�k(qï¼0,kâ 0ä¸kâ 1,a1ï¼0)ï¼ä¸¤è¾¹å常ç¨å¯¹æ°ï¼å¾lgan+1=klgan+lgqï¼ä»¤lgan=bnï¼åæbn+1=kbn+lgqï¼è½¬å为
ç±»åä¸ï¼
ä¾6�å·²ç¥æ°åï½anï½ä¸ï¼a1=2,an+1=an2ï¼æ±anï¼
解ï¼ç±an+1=an2ï¼0ï¼ä¸¤è¾¹å对æ°å¾lgan+1=2lganï¼ä»¤bn=lganåbn+1=2bnï¼å æ¤æ°åï½bnï½æ¯é¦é¡¹ä¸ºb1=lga1=lg2ï¼å
¬æ¯ä¸º2ççæ¯æ°åï¼æ
bn=2n-1lg2=lg22n-1ï¼å³an=22n-1ï¼
(2)âåæ°æ³â转å为
ç±»åä¸ï¼
éæ¨å¼ä¸ºåçå½¢å¼ï¼an+1=(pan+b)/(qan+c)(anâ 0ï¼pqâ 0,pcâ qb)ï¼
è¥b=0ï¼å¾an+1=pan/(qan+c)ï¼å 为anâ 0ï¼æ以两边ååæ°å¾1/an+1=q/p+c/pan,令bn=1/an,åbn+1=(c/p)bn+q/pï¼è½¬å为
ç±»åä¸ï¼
è¥bâ 0ï¼è®¾an+1+x=y(an+x)/qan+cï¼ä¸å·²ç¥éæ¨å¼æ¯è¾æ±å¾xãyï¼ä»¤bn=an+xï¼å¾bn+1=ybn/qan+cï¼è½¬å为b=0çæ
åµï¼
ä¾7�å¨æ°åï½anï½ä¸ï¼å·²ç¥a1=2,an+1=(3an+1)/(an+3)ï¼æ±é项anï¼
解ï¼è®¾an+1+x=y(an+x)/an+3ï¼åan+1=(y-x)an+(y-3)x/an+3ï¼ç»åå·²ç¥éæ¨å¼å¾
y-x=3, æ以
x=1,
y-3=1ï¼ y=4ï¼
åæan+1+1=4(an+1)/an+3ï¼ä»¤bn=an+1ï¼åbn+1=4bn/bn+2ï¼æ±åæ°å¾1/bn+1=1/2�6�11/bn+1/4ï¼å³1/bn+1-1/2=1/2(1/bn-1/2)ï¼
å æ¤æ°åï½1/bn-1/2ï½æ¯é¦é¡¹ä¸º1/b1-1/2=1/a1+1-1/2=-1/6ï¼å
¬æ¯ä¸º1/2ççæ¯æ°åï¼
æ
1/bn-1/2=(-1/6)(1/2)n-1ï¼ä»èå¯æ±å¾anï¼ æ±æ°åçån项åæ¯é«ä¸æ°å¦ãæ°åãä¸ç« çæå¦éç¹ä¹ä¸ï¼è对äºä¸äºéçå·®æ°åï¼åéçæ¯æ°åçæäºæ°åæ±åï¼æ¯ææçé¾ç¹ãä¸è¿ï¼åªè¦è®¤çå»æ¢æ±è¿äºæ°åçç¹ç¹ãåç»æï¼ä¹å¹¶éæ è§å¾å¯å¾ªã
å
¸å示ä¾ï¼
1ã ç¨é项å
¬å¼æ³ï¼
è§å¾ï¼è½ç¨é项å
¬å¼ååºæ°åå项ï¼ä»èå°å
¶åéæ°ç»å为å¯æ±æ°ååã
ä¾1ï¼æ±5ï¼55ï¼555ï¼â¦ï¼çån项åã
解ï¼âµan= 5 9(10n-1)
â´Sn = 5 9(10-1)+ 5 9(102-1) + 5 9(103-1) + ⦠+ 5 9(10n-1)
= 5 9[ï¼10+102+103+â¦+10nï¼-n]
= ï¼10nï¼1-9n-10ï¼
2ã éä½ç¸åæ³ï¼
ä¸è¬å°å½¢å¦{an�6�1bn}çæ°åï¼{ an }为çå·®æ°åï¼ { bn }为çæ¯æ°åï¼åå¯ç¨éä½ç¸åæ³æ±åã
ä¾2ï¼æ±ï¼Sn=1+5x+9x2+�6�1�6�1�6�1�6�1+(4n-3)xn-1
解ï¼Sn=1+5x+9x2+�6�1�6�1�6�1�6�1+(4n-3)xn-1 â
â 两边åä¹ä»¥xï¼å¾
x Sn=x+5 x2+9x3+�6�1�6�1�6�1�6�1+(4n-3)xn â¡
â -â¡å¾ï¼ï¼1-xï¼Sn=1+4ï¼x+ x2+x3+�6�1�6�1�6�1�6�1+ ï¼-ï¼4n-3ï¼xn
å½x=1æ¶ï¼Sn=1+5+9+�6�1�6�1�6�1�6�1+ï¼4n-3ï¼=2n2-n
å½xâ 1æ¶ï¼Sn= 1 1-x [ 4x(1-xn) 1-x +1-ï¼4n-3ï¼xn ]
3ã è£é¡¹æµæ¶æ³ï¼
è¿ä¸ç±»æ°åçç¹å¾æ¯ï¼æ°åå项æ¯çå·®æ°åæç¸é»ä¸¤é¡¹æå 项ç积ï¼
ä¸è¬å°ï¼{an}æ¯å
¬å·®ä¸ºdççå·®æ°åï¼åï¼
å³è£é¡¹æµæ¶æ³ï¼ å¤ç¨äºåæ¯ä¸ºçå·®æ°åçæç¸é»k项ä¹ç§¯ï¼èåå为常éçåå¼åæ°åçæ±åï¼å¯¹è£é¡¹æµæ¶æ³æ±åï¼å
¶è£é¡¹å¯éç¨å¾
å®ç³»æ°æ³ç¡®å®ã
ä¾3ï¼æ± 1 3ï¼ 1 1 5ï¼ 1 3 5ï¼ 1 63ä¹åã
解ï¼
4ã åç»æ³ï¼
æäºæ°åï¼éè¿éå½åç»ï¼å¯å¾åºä¸¤ä¸ªæå 个çå·®æ°åæçæ¯æ°åï¼ä»èå¯å©ç¨çå·®æ°åæçæ¯æ°åçæ±åå
¬å¼åå«æ±åï¼ä»èå¾åºåæ°åä¹åã
ä¾4ï¼æ±æ°å çån项åã
解ï¼
5ã èåæ³ï¼
æçæ°å表示形å¼è¾å¤æï¼æ¯ä¸é¡¹æ¯è¥å¹²ä¸ªæ°çåï¼è¿æ¶å¸¸éç¨èåæ³ï¼
å
对å
¶ç¬¬n项æ±åï¼ç¶åå°é项åç®ï¼ä»èæ¹ååæ°åçå½¢å¼ï¼æå©äºæ¾åºè§£é¢åæ³ã
ä¾5ï¼æ±æ°å2ï¼2+4ï¼2+4+6ï¼2+4+6+8ï¼â¦ï¼2+4+6+â¦+2nï¼â¦çån项å
解ï¼âµan=2+4+6+â¦+2n= n(n+1)=n2+n
â´Sn=(12+1)+(22+2)+(32+3) +â¦â¦+( n2+n)
=ï¼12+22+32+â¦+ n2ï¼+ï¼+2+3+â¦+nï¼
= nï¼n+1ï¼ï¼2n+1ï¼+ n(n+1)
= 1 3nï¼n+1ï¼ï¼n+2ï¼
6ã ååºç¸å æ³ï¼
çå·®æ°åån项åå
¬å¼çæ¨å¯¼ï¼æ¯å
å°åå¼ä¸å项ååºç¼æå¾åºå¦ä¸ä¸ªåå¼ï¼ç¶ååä¸åæ¥çåå¼å¯¹åºç¸å ï¼ä»è解å¾çå·®æ°åçån项åå
¬å¼ï¼å©ç¨è¿ç§æ¹æ³ä¹å¯ä»¥æ±åºæäºæ°åçån项åã
ä¾6ï¼å·²ç¥lg(xy)=aï¼æ±Sï¼å
¶ä¸
S=
解: å°åå¼Sä¸å项ååºæåï¼å¾
å°æ¤åå¼ä¸ååå¼ä¸¤è¾¹å¯¹åºç¸å ï¼å¾
2S= + + �6�1 �6�1 �6�1 +
(n+1)项
=n(n+1)lg(xy)
âµ lg(xy)=a â´ S= n(n+1)a
以ä¸ä¸ä¸ª6ç§æ¹æ³è½ç¶åæå
¶ç¹ç¹ï¼ä½æ»çååæ¯è¦åäºæ¹ååæ°åçå½¢å¼ç»æï¼ä½¿å
¶è½è¿è¡æ¶é¡¹å¤çæè½ä½¿ç¨çå·®æ°åæçæ¯æ°åçæ±åå
¬å¼ä»¥åå
¶å®å·²ç¥çåºæ¬æ±åå
¬å¼æ¥è§£å³ï¼åªè¦å¾å¥½å°ææ¡è¿ä¸è§å¾ï¼å°±è½ä½¿æ°åæ±ååé¾ä¸ºæï¼è¿åè解ã
温馨提示:答案为网友推荐,仅供参考