when finding consecutive odd integers why would you use n, n+2 and n-2 rather than n, n+2 and n+4
you should use neither...you should use 2n+1,2n+3, and 2n+5 to get odd integers
you could use either,
(n-2) , n , (n+2) or
n , n+2, n+4 or
2n+1, 2n+3, 2n+5 or
12n+34. 12n+36 , 12n+38 or
...
since odd numbers are 2 apart, any of the above would work.
the advantage of using n-2, n, and n+2 is that when they are added, the constants would drop out
thanks
When finding consecutive odd integers, you can certainly use either of the two approaches you mentioned. Both n, n+2, and n+4 as well as n, n+2, and n-2 can be used to represent consecutive odd integers. The choice between the two options depends on the specific problem or context in which you are working.
If you choose to use n, n+2, and n+4, it means you are representing three consecutive odd integers with an increment of 2 between each number. For example, if you assign n a value of 3, then n+2 would be 5, and n+4 would be 7. Similarly, if n is 9, then the three consecutive odd integers would be 9, 11, and 13.
On the other hand, if you use n, n+2, and n-2, you are representing three consecutive odd integers with an increment of 2 between the first and second number, and a decrement of 2 between the second and third number. In this case, if n is assigned a value of 3, then n+2 would be 5, and n-2 would be 1. Therefore, the consecutive odd integers would be 1, 3, and 5.
Both representations are valid; however, the choice depends on the specific problem or equation you are working with. Consider the given information or constraints of the problem and decide which approach best fits the situation.