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Can sum of the first few consecutive natural numbers is 500? - Answers

No.The sum of the first n consecutive natural numbers is:sum = n(n+1)/2So if the sum is 500, then:n(n+1)/2 = 500⇒ n(n+1) = 1000Which means that 1000 must have a pair of factors with a difference of 1 if it is possible.The pairs factors of 1000 are:1 x 10002 x 5004 x 2505 x 2008 x 12510 x 10020 x 5025 x 40No two are separated by 1, thus the sum is impossible.



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Can sum of the first few consecutive natural numbers is 500? - Answers

https://math.answers.com/math-and-arithmetic/Can_sum_of_the_first_few_consecutive_natural_numbers_is_500

No.The sum of the first n consecutive natural numbers is:sum = n(n+1)/2So if the sum is 500, then:n(n+1)/2 = 500⇒ n(n+1) = 1000Which means that 1000 must have a pair of factors with a difference of 1 if it is possible.The pairs factors of 1000 are:1 x 10002 x 5004 x 2505 x 2008 x 12510 x 10020 x 5025 x 40No two are separated by 1, thus the sum is impossible.



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https://math.answers.com/math-and-arithmetic/Can_sum_of_the_first_few_consecutive_natural_numbers_is_500

Can sum of the first few consecutive natural numbers is 500? - Answers

No.The sum of the first n consecutive natural numbers is:sum = n(n+1)/2So if the sum is 500, then:n(n+1)/2 = 500⇒ n(n+1) = 1000Which means that 1000 must have a pair of factors with a difference of 1 if it is possible.The pairs factors of 1000 are:1 x 10002 x 5004 x 2505 x 2008 x 12510 x 10020 x 5025 x 40No two are separated by 1, thus the sum is impossible.

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      No.The sum of the first n consecutive natural numbers is:sum = n(n+1)/2So if the sum is 500, then:n(n+1)/2 = 500⇒ n(n+1) = 1000Which means that 1000 must have a pair of factors with a difference of 1 if it is possible.The pairs factors of 1000 are:1 x 10002 x 5004 x 2505 x 2008 x 12510 x 10020 x 5025 x 40No two are separated by 1, thus the sum is impossible.

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