正整數 1,2,3 ..... 依下列方法分成為 G1 , G2 , G3 ....各組 , 使第k組Gk 由K個連續組成
G1: 1
G2: 2,3
G3: 4,5,6
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Gk-1: u1 , u2 , ..... , uk-1
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a) 1.2 part唔關事
b. Find , in terms of k
1)the last integer uk-1 in Gk-1
ans : [1+(k-1)](k-1)/2
我想問既係佢要搵 k-1 term 果整數~~定係加埋既整數
點解會用左 n/2[2a+(n-1)d]<<<呢公式
A.S 同G.S 一個問題
2008-09-29 7:41 am
回答 (1)
2008-10-04 5:48 pm
✔ 最佳答案
For G(1). last term is 1.For G(2), last term is 3 = 1 + 2
For G(3), last term is 6 = 1 + 2 + 3
For G(4): 7, 8, 9, 10. Last term is 10 = 1 + 2 + 3 + 4
For G(5): 11, 12, 13, 14, 15. Last term is 15 = 1 + 2 + 3 + 4 + 5 and so on.
For G(n): Last term will be 1 + 2 + 3 + 4 + 5 +...... + n = S
Using the formula S = n[2a + (n-1)d]/2. Now a = 1, d = 1, so S = n(n+1)/2.
That is the last term of G(n) is n(n+1)/2.
Therefore, the last term of G(k - 1) = u(k- 1) = (k -1)[(k-1) + 1]/2.
收錄日期: 2021-04-25 22:37:24
原文連結 [永久失效]:
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