標題:

我想問下一條通項

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發問:

我想問: (A) 3,9,19,33,51... (B) 3,10,29,66,127... 更新: (C) 3,10,21,36,55..... (D) 4,12,24,40,60.....

最佳解答:

(A) 3,9,19,33,51,73,99 (B) 3,10,29,66,127,218,345,514,731,1002,1333,1730,........ 2006-11-04 05:01:30 補充: c) 3,10,21,36,55,78....(n-1)既二次方 (n)x(n-1)(n-1)(n-1)即是1既二次方 2x1 = 32既二次方 3x2 = 103既二次方 4x3 = 214既二次方 5x4 = 365既二次方 6x5 = 55from http://hk.knowledge.yahoo.com/question/?qid=7006092800815 2006-11-04 05:07:42 補充: d) 4,12,24,40,60, 84, 112, 144, 180 from : 2006-11-04 05:08:27 補充: Explaination for D.#1 = 4 = 4 #2 = 4 8 = 12 #3 = 4 8 12 = 24 #4 = 4 8 12 16 = 40 #5 = 4 8 12 16 20 = 60 The final # is equal to half the last number being added times (n 1). For example, in #3, half of 12 times 4 = 24. Or in #4, half of 16 times 5 = 40.

其他解答:

(A) 73 3+6=9 9+10=19 19+14=33 33+18=51 51+22=73 The differences will be 6,10,14,18,22,26,30,34...X+4 (B) 218 3+7=10 10+19=29 29+37=66 66+61=127 127+91=218 The differences will be 7,19,37,61,91....(X(n-1)-X(n-2)+6)|||||(a) general term= 2n^2 + 1 (it is transformed from the sequence of square number) (b) general term= n^3 + 2(derived from cubic number) 2006-11-02 15:49:38 補充: Both c and d derived from triangular no.(c) 2n2 n(d) 2n2 2n 2006-11-02 15:50:52 補充: there are problems in display.(c) 2n^2 n(d) 2n^2 2n 2006-11-02 15:52:06 補充: (c) 2n^2 + n(d) 2n^2 + 2n
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