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標題:
general term~10,6,3,1
發問:
10,6,3,1 ge general term係咩? 我知1,3,6,10 係三角形數 so general term is (n+1)x n/2 but this 10,6,3,1 is倒三角形 how to find it?(中文)
最佳解答:
T(1) = 10 T(2) = 10 - 4 T(3) = 10 - (4 + 3) T(4) = 10 - (4 + 3 + 2) ..............................T(n) = 10 - (4 + 3 + 2 + 1 + 0 + -1 + -2 + ...)括弧內有 n-1 項。等差數列和 4 + 3 + 2 + 1 + 0 + -1 + -2 + ... 首項 = 4 公差 d = - 1 項數 = n - 1故尾項 = 4 + (n-2)(- 1) = 6 - n 4 + 3 + 2 + 1 + 0 + -1 + -2 + ... (n-1 項) = (首項 + 尾項) x 項數 / 2 = (4 + 6-n)(n - 1) / 2 = (10 - n)(n - 1) / 2即 T(n) = 10 - (10 - n)(n - 1) / 2= [20 + (n - 10)(n - 1)] / 2= (n2 - 11n + 30) / 2= (n - 5) (n - 6) / 2
其他解答:
general term~10,6,3,1
發問:
10,6,3,1 ge general term係咩? 我知1,3,6,10 係三角形數 so general term is (n+1)x n/2 but this 10,6,3,1 is倒三角形 how to find it?(中文)
最佳解答:
T(1) = 10 T(2) = 10 - 4 T(3) = 10 - (4 + 3) T(4) = 10 - (4 + 3 + 2) ..............................T(n) = 10 - (4 + 3 + 2 + 1 + 0 + -1 + -2 + ...)括弧內有 n-1 項。等差數列和 4 + 3 + 2 + 1 + 0 + -1 + -2 + ... 首項 = 4 公差 d = - 1 項數 = n - 1故尾項 = 4 + (n-2)(- 1) = 6 - n 4 + 3 + 2 + 1 + 0 + -1 + -2 + ... (n-1 項) = (首項 + 尾項) x 項數 / 2 = (4 + 6-n)(n - 1) / 2 = (10 - n)(n - 1) / 2即 T(n) = 10 - (10 - n)(n - 1) / 2= [20 + (n - 10)(n - 1)] / 2= (n2 - 11n + 30) / 2= (n - 5) (n - 6) / 2
其他解答:
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