This kind of question is frequently asked and easily solved. 1 + 1000 = 1001 2 + 999 = 1001 3 + 998 = 1001 You can see the developing pattern here. Taking integers in order from the top and bottom of the list allows you to create hundreds of equal sums. The last such sum that can be made is 500 + 5001 = 1001. It is clear that you then have 500 sums, each equal to 1001. 1001 X 500 = 500,500 which is the sum of all the integers combined.
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Using Gauss's method, 1+2+3...1000= 500x1001=500500 Answer:500500
No.Let the four numbers be (n-1), n, (n+1), (n+2).Their sum is 4n + 2 = 2(2n + 1)If 2000 is the sum of four consecutive integers, then:2(2n+1) = 2000⇒ 2n + 1 = 1000⇒ 2n = 999but 999 is odd, not even and so n cannot be an integer; therefore 2000 is not the sum of four consecutive integers.
(300 x (300 + 1)) / 2 = 45150 Therefore, the sum of all the integers from 1 to 300 is 45150.
The first odd positive integers are "1" and "3" which the sum is 4.
To find the number of even integers between 100 and 1000, we first determine the number of even integers between 1 and 1000, which is half of the total integers (since every other integer is even). So, 1000/2 = 500 even integers between 1 and 1000. Next, we subtract the number of even integers between 1 and 100, which is 50 (since every other integer is even in this range as well). Therefore, there are 500 - 50 = 450 even integers between 100 and 1000.