項數計算方法

求項數:
[ (頭-尾)/分距 ] +1
= [ (2004-6) / 6 ] +1
= 333+1
= 334

每3個數字group埋1堆:
2004 + (1998 - 1992) = 2010
1986 + (1980 - 1974) = 1992
......

334個項數, 咁就應該有334/3 = 111個group, 剩返個+6
每個group應該相差18....
(因為括號中所得出既數字應該係1樣
但2004同1986相差18, 1986又應該同下1個group數字既第1個數字相差18...
如此類推)

咁條式即係:
2010 + 1992 + (1992-18) + ...... [total加111個數]

( 頭+尾 ) *項數 除2
= 2010+[2010-(110*18)] *項數 除2
= (2010+30)*111 除2
=113220

再加返剩左果個+6

答案 = 113226

2004+1998-1992+1986+1980-1974+.....+6
=6x(334+333-332+331+330-329+328+327-326+...+4+3-2+1)
=6x(334+333+332+331+330+329+328+327+326+...+1)-12x(332+329+326+323+...+2)

first part being an AP of common diff 1, first term 1 and total 334 terms
second part being an AP of common diff 3, first term 2 and total 111 terms
therefore,
amount=6x334x335/2-12x111/2x(2+332)
=335670-222444
=113226

試看看2004+1998-1992=2004+6=(18x111)+6+6=1(8x111)+12
再看看1986+1980-1974=1986+6=(18x110)+6+6=(8x110)+12
如此類推
直到24+18-12=24+6=(18x1)+12
-12之後是+6

所以全式可寫成
(18)(111+110+......+1)+(12x111)+6
=18[(111+1)(111)/2] +1332+6
=18x56x111+1338
=112x999+1338
=112x1000+1338-112
=112000+1226
=113226

2006-12-05 22:48:19 補充：
試看看2004+1998-1992=2004+6=(18x111)+6+6=(18x111)+12再看看1986+1980-1974=1986+6=(18x110)+6+6=(18x110)+12這樣才對

2004/6
= 334

由1加到334...
1+2+3+...+333+334
= [(1+334)*334]/2
= 55945

a = 2004
d = 1998-2004 = -6
a+nd = 6
2004+n(-6) = 6
n = 333

( T(1) + T(333) )* 333 /2
(2004+6)*333/2 =334 665