The consecutive numbers are numbers such that n2 - n1 = 1, where the n's are preceding number in the sequence. In other words, n2 follows after n1.
Consider this example:
1, 2, 3 and 4
Then, 1 + 2 + 3 + 4 = 10
How? 1 + 3 is even and 2 + 4 is even. Then, the sum of two even integers is even.
So referring to the example we have, we can see that the sum of two (2k - 1)-th numbers is even. The sum of two 2k-th numbers is even. Combine these sums altogether to obtain the even number.
Make note that the consecutive numbers don't have to start with an odd integer, such as 1. For instance:
Add up 2, 3, 4 and 5, and we obtain even number, which is 14 = 2 + 3 + 4 + 5.
"Symbolically" or simply:
n1 + n3 = 2k where k is any integer
n2 + n4 = 2l where l is any integer
"2k" and "2l" means even sums.
Combine these values altogether to obtain:
(n1 + n3) + (n2 + n4)
= 2k + 2l
= 2(k + l) where k + l belongs to the set of integers
So we obtain even sum of any four consecutive counting numbers.
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No. The sum of four consecutive integers is two odd numbers plus two even numbers which is an even number. 2001 is an odd number, therefore it cannot be the sum of four consecutive numbers.
There is no set of four consecutive numbers (odd or even) whose sum equals 169.
The numbers are 28, 30, 32 and 34.
The numbers are 82, 84, 86 and 88.
There is no set of four consecutive numbers with a product of 182. There is a set of four consecutive numbers with a sumof 182: 9, 20, 21 and 22.