小学奥数拆分:2*3/(1*4)+5*6/(4*7)+8*9/(7*10)-……+98*99/(97*100) 2*3/1*4+5*6/4*7+8*9/7*10+。。。。。...
\u5c0f\u5b66\u5965\u6570\u62c6\u5206\uff1a1/(1*2*3*4)-1/(2*3*4*5)-1/(3*4*5*6)-\u2026\u2026-1/(6*7*8*9)-1/(7*8*9*10)\u770b\u56fe\u7247\u662f+\u4e0d\u662f-\u554a\uff1f
\u5148\u6309+\u7b97\u3002
\u9996\u5148
1/[n*(n+1)*(n+2)*(n+3)]=1/3 * {1/[n*(n+1)*(n+2)]-1/[(n+1)*(n+2)*(n+3)]}\u6240\u4ee5\uff0c\u539f\u5f0f=1/3[(1/1*2*3-1/2*3*4)+(1/2*3*4-1/3*4*5)...+(1/7*8*9-1/8*9*10)]=1/3(1/1*2*3-1/8*9*10)=119/2160
\u3000\u3000
原式=1+2/(1*4) +1+2*(4*7) + 1 + 2/ (7*10) +.....+ 1+ 2/(97*100)
根据等差数列: 97= 1 + (n-1) *3 所以求出 这个项数一共有33项,所以
原式=33 + 2/3 * [ 1- 1/4 +1/4 - 1/7 + 1/7 - 1/10 +......+ 1/97 - 1/100) ]
= 33 + 2/3 * ( 1- 1/100) = 33 +33/ 50=1683/ 50
解答完毕
第一项为100*98*97
[1/(99*98*97)]
第二项为100*98*97*[1/(98*97*96)]
依次类推,最后一项为
100*99*98*[1/(3*2*1)]
(总共97项)
现在只需要计算1/(99*98*97)
+。。。。。+1/(3*2*1)
1/(99*98*97)=
(99*98+97*98-
2*99*97)
/(2*99*98*97)=1/(2*97)+1/(2*99)
-
1/98
1/(99*98*97)
+。。。。。+1/(3*2*1)
=(
1/2*97+1/2*96+....+1/2*3)+
(1/2*99+1/2*98+....+1/2*5)
-
(1/98+1/97+.......1/4)
所有1/2*97
到
1/2*5的可以全部消掉
=1/2*4
+1/2*3
+
1/2*99+1/2*98
-
1/98
-
1/4
后面的自己验证结果
答案是2
2
绛旓細瑙f瀽锛氾紙a+1锛*锛坅+2锛/a*锛坅+3锛=1+2/3 *锛1/a-1/a+3锛夛紙瑁傚樊鍏紡鎺ㄧ悊鐪佺暐锛夊師寮=1+2/3*锛1/1-1/4锛+1+2/3*锛1/4-1/7锛+...+1+2/3*锛1/97-1/100锛=33+2/3*锛1-1/100锛=33鍙33/50
绛旓細鍘熷紡锛1锛2/(1*4) +1+2*(4*7) + 1 + 2/ (7*10) +...+ 1+ 2/(97*100)鏍规嵁绛夊樊鏁板垪锛氥97= 1 + (n-1) *3 鎵浠ユ眰鍑恒杩欎釜椤规暟涓鍏辨湁33椤癸紝鎵浠 鍘熷紡锛33 + 2/3 * [ 1- 1/4 +1/4 - 1/7 + 1/7 - 1/10 +...+ 1/97 - 1/100) ]= 33 + 2/3...
绛旓細鍘熷紡=锛3-1锛/2*3+(4-1)/(2*3*4)+(5-1)/(2*3*4*5)+(6-1)/(2*3*4*5*6)+(7-1)/(2*3*4*5*6*7)=[3/2*3+4/(2*3*4)+5/(2*3*4*5)+6/(2*3*4*5*6)+7/(2*3*4*5*6*7)]-[1/2*3+1/(2*3*4)+1/(2*3*4*5)+1/(2*3*4*5*6)+1/(2...
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绛旓細...99*100=(1/3)*[99*100*101-98*99*100]1*2+2*3+3*4+鈥︹99*100 =(1/3)*[99*100*101]=333300 =1*(1+1)+2*(2+1)+3*(3+1)+...99*(99+1)=1鐨勫钩鏂+2鐨勫钩鏂+3鐨勫钩鏂+...99鏂+1+2+3+...99 =1/6*99*(99+1)*(2*99+1)+(1+99)99/2 =333300 ...
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绛旓細鍥犱负锛 1 2=2*3/2 1 2 3=3*4/2 1 2 3 4=4*5/2 1 2 3 鈥︹ 2006=2006*2007/2 鎵浠ワ紝 1 1/(1 2) 1/(1 2 3) 1/(1 2 3 4) ... 1/(1 2 3 ... 2006) =1 2/锛2*3锛 2/锛3*4锛 2/锛4*5锛 鈥︹ 2/锛2006*2007锛 =2[锛1/2 1/锛2*3锛 1/锛3...
绛旓細鐪嬪浘鐗囨槸+涓嶆槸-鍟婏紵鍏堟寜+绠椼傞鍏 1/[n*(n+1)*(n+2)*(n+3)]=1/3 * {1/[n*(n+1)*(n+2)]-1/[(n+1)*(n+2)*(n+3)]}鎵浠ワ紝鍘熷紡=1/3[(1/1*2*3-1/2*3*4)+(1/2*3*4-1/3*4*5)...+(1/7*8*9-1/8*9*10)]=1/3(1/1*2*3-1/8*9*10)=119/...
绛旓細瑙o細鏈鏈変笁涓В锛屼竴鍏辨湁7鍙洅瀛愶紝4鍙洅瀛愶紝3鍙洅瀛愩傜偣閲戞湳锛氬阀鐢ㄥ亣璁惧拰鎺ㄧ悊鎶婂凡鐭ュ拰鏈煡鑱旂郴璧锋潵銆備緥2銆佸皢1992琛ㄧず鎴愯嫢骞蹭釜鑷劧鏁扮殑鍜,濡傛灉瑕佷娇杩欎簺鏁扮殑涔樼Н,杩欎簺鑷劧鏁版槸___.(1992骞存姹夊競灏忓鏁板绔炶禌璇曢)璁叉瀽:鑻ユ妸涓涓暣鏁鎷嗗垎鎴愬嚑涓嚜鐒舵暟鏃,鏈夊ぇ浜4鐨勬暟,鍒欐妸澶т簬4鐨勮繖涓暟鍐嶅垎鎴愪竴...
绛旓細瑙o細1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+鈥︹+1/(98*99*100)=(1/2)*(4-3)/(3*4)+(1/3)*(5-4)/(4*5)+(1/4)*(6-5)/(5*6)+鈥︹+(1/98)*(100-99)*(99*100)=(1/2)*(1/3-1/4)+(1/3)*(1/4-1/5)+(1/4)*(1/5-1/6)+鈥︹+(1/98)*(1/...