Discussion
If n > 2, then the sum, S, of the integers from 1 through n can be calculated by the following formula: S = n(n + 1)/2. Which one of the following statements about S must be true?
*This question is included in Nova Math - Problem Set F: Number Theory, question #22
| (A) | S is always odd. |
| (B) | ... |
| (C) | ... |
| (D) | ... |
| (E) | ... |
| (F) | ... |
The solution is
Posted: 10/05/2012 13:23
Isn't answer B also correct? Since you are multiplying n and (n+1), you will always be multiplying an odd and an even, and therefore the product should always be even, which when divided by 2 equals an even number. Thanks for your help!
Posted: 10/05/2012 14:20
Kayla, you are right that the product of n and n+1 will always be even. But, an even number divided by 2 is not always even. 10 divided by 2 is 5, which is not even.
Posted: 10/05/2012 14:31
I see now. I greatly appreciate your help.
Posted: 02/03/2014 13:25
Are you not multiplying consecutive integers greater than 2? n and n+1 would be consecutive, right? So your response using the example of 10/2 equals 5 does not make sense to me. If n could be equal to 2 then it would be 6/2=3 and your example would make sense, but n is greater than 2. The integers greater than 2 that I plugged in produced an even S, so I don't understand why B is not an answer as well.
Posted: 02/04/2014 10:51
Lauren, a very good question. You are right that n (n+1) will always be even.
Here is a better example:
n=5, n+1=6. 5x6=30.
S=n*(n+1)/2 = 30/2 = 15. Not even.
Another one: n=6. S=6x7/2 = 21. Not even.
n=9. S=90/2 = 45. Not even. And so on.
Here is a better example:
n=5, n+1=6. 5x6=30.
S=n*(n+1)/2 = 30/2 = 15. Not even.
Another one: n=6. S=6x7/2 = 21. Not even.
n=9. S=90/2 = 45. Not even. And so on.