Is 1+2+3+4+...+(n-1) divisible by N (odd positive integer)?

[Isaak DeMaio]

That is what I did, though I didn't use induction to prove 1+2+3+...+k=k*(k+1)/2, I just sort of took that for granted. Then it is pretty easy to show what you need to show. But, it seems that Mark44 has done this using a direct inductive argument, is that correct Mark44? I haven't come up with one, but I haven't really given it much thought and I doubt I will.

where do you get this divided by 2 from?

 

[Isaak DeMaio]

Ok, nothing is being divided by K, I'm not sure what you are referring to, perhaps I made a typo somewhere.

if 1 + 2 + ... + k = k(k+1)/2, what can you conclude?

I don't understand where you are getting a 2 from.

 

[Isaak DeMaio]

Base case:
1 + 2 = 3 0 (mod 3)
This establishes the base case.

Induction step:
Let n = k
Assume that 1 + 2 + 3 + ... + (k - 1) = 0 (mod k) is true. (*)

Now let n = k + 2 (to get to the next odd number)
Show that 1 + 2 + 3 + ... + (k - 1) + k + (k + 1) = 0 (mod k + 2).

Where do you go a denominator of 2 for any of this?

 

[Robert1986]

Base case:
1 + 2 = 3 0 (mod 3)
This establishes the base case.

Induction step:
Let n = k
Assume that 1 + 2 + 3 + ... + (k - 1) = 0 (mod k) is true. (*)

Now let n = k + 2 (to get to the next odd number)
Show that 1 + 2 + 3 + ... + (k - 1) + k + (k + 1) = 0 (mod k + 2).

Where do you go a denominator of 2 for any of this?

So, here is how I would prove this, which I think departs slightly from what Mark44 was doing:

1)For any integer k, 1+2+...+k=k(k+1)/2, prove this using induction. If you have proven things via induction as you say you have, this is a piece of cake.

2)Use this to conclude that if k+1 is odd, then 1+2+...+k is divisible by k+1, that is 1+2+...+k = 0 (mod k+1)

I've just about spelled it out as clearly as I can without giving you the answer, go for it, and don't respond for at least half an hour; think about it before responding.

 

[Isaak DeMaio]

So, here is how I would prove this, which I think departs slightly from what Mark44 was doing:

1)For any integer k, 1+2+...+k=k(k+1)/2, prove this using induction. If you have proven things via induction as you say you have, this is a piece of cake.

2)Use this to conclude that if k+1 is odd, then 1+2+...+k is divisible by k+1, that is 1+2+...+k = 0 (mod k+1)

I've just about spelled it out as clearly as I can without giving you the answer, go for it, and don't respond for at least half an hour; think about it before responding.

Say this again but in induction form so I can understand it.

 

[Robert1986]

Say this again but in induction form so I can understand it.

No. You didn't even wait ten minutes. Did you think about it? Look, if you expect to do well in Math, you can't expect that you will always have someone over your shoulder telling you what to do next. I am not going to write this in "induction form", but I will reiterate what I said before:

1)Prove that for any integer k, 1+2+...+k = k(k+1)/2. You have said that you can do inductive proofs, this is a rather basic inductive proof, in fact, it is the most basic inductive proof I can think of. If you can do induction, you can do this. And I am not going to tell you how to use induction, you are going to have to figure that out own your own. Spend a little time, put some effort into it.

2)So now that you know that 1+2+...+k=k(k+1)/2 what can you conclude? This part involves NO INDUCTION AT ALL. Think about these things: Is k even or odd? Is k+1 even or odd? Is k/2 an integer? Is (k+1)/2 an integer? How do the answers to these questions relate to what I am trying to prove?


Now, spend some time on this. You are giving up way too easily. Like I said, if you want to be successful at math, whether you are doing research, working in some applied area or teaching, you are going to have to figure stuff out for yourself. We have given you ample "hints" (we have, in fact, done much more than that), if you refuse to think about it, there really isn't a lot more that we can do.

 

[Isaak DeMaio]

No. You didn't even wait ten minutes. Did you think about it? Look, if you expect to do well in Math, you can't expect that you will always have someone over your shoulder telling you what to do next. I am not going to write this in "induction form", but I will reiterate what I said before:

1)Prove that for any integer k, 1+2+...+k = k(k+1)/2. You have said that you can do inductive proofs, this is a rather basic inductive proof, in fact, it is the most basic inductive proof I can think of. If you can do induction, you can do this. And I am not going to tell you how to use induction, you are going to have to figure that out own your own. Spend a little time, put some effort into it.

2)So now that you know that 1+2+...+k=k(k+1)/2 what can you conclude? This part involves NO INDUCTION AT ALL. Think about these things: Is k even or odd? Is k+1 even or odd? Is k/2 an integer? Is (k+1)/2 an integer? How do the answers to these questions relate to what I am trying to prove?


Now, spend some time on this. You are giving up way too easily. Like I said, if you want to be successful at math, whether you are doing research, working in some applied area or teaching, you are going to have to figure stuff out for yourself. We have given you ample "hints" (we have, in fact, done much more than that), if you refuse to think about it, there really isn't a lot more that we can do.

It's an extra credit question it's obviously not as easy you say than the 5 hours i spent on the rest of the test.

 

[berkeman]

Unfortunately, Isaak is no longer with us. Thank you all for trying so hard to help him. Thread closed.