Maths Question--------- 24
2007-04-23 6:12 am
Find the sum of the first twenty odd positive integers that have a remainder of 1 when divided by 3 .
回答 (3)
2007-04-23 7:53 am
✔ 最佳答案
3n + 1 = a positive odd number3n = a positive even number or 0
n = a positive even number or 0
therefore
The sum of the first twenty odd positive integers that have a remainder of 1 when divided by 3
= 3 x (0 + 2 + 4 + 6 + ... + 38) + 1 x 20
= 3 x (0 + 38) x 20 x 1/2 + 20
= 3 x 38 x 10 + 20
= 1160
2007-04-23 6:43 am
n=12
a=1
d=3
12/2[2(1)+(12-1)(3)]
=210_______
a=1
d=3
12/2[2(1)+(12-1)(3)]
=210_______
2007-04-23 6:28 am
Isn't it 400? I think you only add 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39~because they are the first twenty odd positive integers~and it does give a remainder of 1 when divided by 3.
參考: myself
收錄日期: 2021-04-30 21:01:19
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