標題:
two easy maths!!
發問:
given that the sum of the integers from 1 to n is n(n+1)/2 (a)find 51+52+53+..............+100 (b)find 21+22+23+....+50 更新: 我要條式呀-3-
1 + 2 + 3 + ...... + n = n(n + 1)/2 (a) 51 + 52 + 53 + ...... + 100 = [1 + 2 + 3 + ...... + 100] - [1 + 2 + 3 + ...... + 50] = [100(100 + 1)/2] - [50(50 + 1)/2] = 5050 - 1275 = 3775 ===== (b) 21 + 22 + 23 + ...... + 50 = [1 + 2 + 3 + ...... + 50] - [1 + 2 + 3 + ...... 20] = [50(50 + 1)/2] - [20/(20 + 1)/2] = 1275 - 210 = 1065 =
其他解答:
a) (1+100)100/2-(1+50)50/2 =5050-1275 =3775 b) (1+50)50/2-(1+20)20/2 =1275-210 =1065|||||a:(51+100)x50÷2 151x50÷2 151x25 =3775 b:(21+50)x30÷2 71x30÷2 71x15 =1065 2008-10-11 16:54:05 補充: 答案得來的方法是: (頭項+未項)x項數÷2 ( 51 + 100) x 50 ÷ 2 151x50÷2 151x25 =3775 項數得來的方法是: (未項-頭項)+1 ( 100 - 51 )+1 =49+1 =50|||||a = 3775 b = 1065
two easy maths!!
發問:
given that the sum of the integers from 1 to n is n(n+1)/2 (a)find 51+52+53+..............+100 (b)find 21+22+23+....+50 更新: 我要條式呀-3-
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最佳解答:1 + 2 + 3 + ...... + n = n(n + 1)/2 (a) 51 + 52 + 53 + ...... + 100 = [1 + 2 + 3 + ...... + 100] - [1 + 2 + 3 + ...... + 50] = [100(100 + 1)/2] - [50(50 + 1)/2] = 5050 - 1275 = 3775 ===== (b) 21 + 22 + 23 + ...... + 50 = [1 + 2 + 3 + ...... + 50] - [1 + 2 + 3 + ...... 20] = [50(50 + 1)/2] - [20/(20 + 1)/2] = 1275 - 210 = 1065 =
其他解答:
a) (1+100)100/2-(1+50)50/2 =5050-1275 =3775 b) (1+50)50/2-(1+20)20/2 =1275-210 =1065|||||a:(51+100)x50÷2 151x50÷2 151x25 =3775 b:(21+50)x30÷2 71x30÷2 71x15 =1065 2008-10-11 16:54:05 補充: 答案得來的方法是: (頭項+未項)x項數÷2 ( 51 + 100) x 50 ÷ 2 151x50÷2 151x25 =3775 項數得來的方法是: (未項-頭項)+1 ( 100 - 51 )+1 =49+1 =50|||||a = 3775 b = 1065
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