# consecutive odd integers

• Mar 28th 2010, 12:59 PM
mjoshua
consecutive odd integers
The sum of 3 consecutive odd integers is k. In terms of k, what is the sum of the 2 smaller of these integers?

Do I need to use substitution somehow?? (The answer given is $\frac {2k}{3} - 2$ but can't figure it out..)
• Mar 28th 2010, 01:06 PM
mr fantastic
Quote:

Originally Posted by mjoshua
The sum of 3 consecutive odd integers is k. In terms of k, what is the sum of the 2 smaller of these integers?

Do I need to use substitution somehow?? (I know it's gonna be $\frac {2k}{3} - 2$ but can't figure it out..)

k - (largest odd integer).

If you post the whole question you might get an answer that's more suitable for your purpose.
• Mar 28th 2010, 01:07 PM
e^(i*pi)
Quote:

Originally Posted by mjoshua
The sum of 3 consecutive odd integers is k. In terms of k, what is the sum of the 2 smaller of these integers?

Do I need to use substitution somehow?? (I know it's gonna be $\frac {2k}{3} - 2$ but can't figure it out..)

Let n be the middle odd integer so we have

$k - n+2$

as Mr F said, you'll need to expand
• Mar 28th 2010, 01:14 PM
mjoshua
That IS the entire question (it's taken out of my book as is) :(
• Mar 28th 2010, 01:30 PM
e^(i*pi)
Quote:

Originally Posted by mjoshua
The sum of 3 consecutive odd integers is k. In terms of k, what is the sum of the 2 smaller of these integers?

Do I need to use substitution somehow?? (The answer given is $\frac {2k}{3} - 2$ but can't figure it out..)

Perhaps the idea of an arithmetic sequence can help us.

Let $n$ be the middle number so that our sequence is $(n-2)+n+(n+2) = k$

For 3 consecutive odd numbers the mean will be equal to the middle number. Hence

$n = \frac{k}{3}$ is the middle number.

Hence the smaller number is: $n-2 = \frac{k}{3} - 2$.

The question is asking us to find the sum of the two smaller numbers:

$(n-2)+n = \left(\frac{k}{3} - 2 \right) + \frac{k}{3} = \frac{2k}{3} - 2$

• Mar 28th 2010, 10:04 PM
mjoshua
Quote:

Originally Posted by e^(i*pi)
Perhaps the idea of an arithmetic sequence can help us.

Let $n$ be the middle number so that our sequence is $(n-2)+n+(n+2) = k$

Why can we use this sequence though? (Should it be a consecutive integer odd sequence? If you try n = 0, or n = 2 it's not completely odd.
• Mar 29th 2010, 11:09 AM
e^(i*pi)
Quote:

Originally Posted by mjoshua
Why can we use this sequence though? (Should it be a consecutive integer odd sequence? If you try n = 0, or n = 2 it's not completely odd.

Neither 0 nor 2 are odd integers.

If you think about an arithmetic sequence it is defined as there being a common difference between each term. For consecutive odd integers this is 2. n is 2 more than n-2 and n+2 is 2 more than n.

If we try $n=3$ and hence $k = 1+3+5 = 9$

$(3-2)+3 = \frac{2(9)}{3} - 2$ and sure enough $4=4$
• Mar 29th 2010, 09:47 PM
mjoshua
But instead of declaring "n" just an odd integer, shouldn't we use a sequence like 2n + 1 so that no matter what you put it for n (even or odd) you still get an odd integer? (so it's more inclusive?)
• Mar 30th 2010, 02:22 AM
mr fantastic
Quote:

Originally Posted by mjoshua
But instead of declaring "n" just an odd integer, shouldn't we use a sequence like 2n + 1 so that no matter what you put it for n (even or odd) you still get an odd integer? (so it's more inclusive?)

You have been given the solution on a platter in post #5. Now it's your job to take some ownership of that solution and modify it if you want to express the odd numbers in the way you describe.
• Mar 30th 2010, 04:49 AM
bjhopper
sum of odd integers
Quote:

Originally Posted by mjoshua
But instead of declaring "n" just an odd integer, shouldn't we use a sequence like 2n + 1 so that no matter what you put it for n (even or odd) you still get an odd integer? (so it's more inclusive?)

Hello Mjoshua,
I was going to reply to this post yesterday but decided that eipi had adequately answered it I passed.Today you appear confused.Here is the way I would have replied.

Three cons odd integers are n, n+2, n+4, the sum of these =k
Solution of this equation n=k-6/3. The sum of the first two is 2n+2 and thisbecomes 2/3k-2 when you insert the value of n in terms of k.

bjh
• Mar 30th 2010, 10:01 AM
mjoshua
Quote:

Originally Posted by mr fantastic
You have been given the solution on a platter in post #5. Now it's your job to take some ownership of that solution and modify it if you want to express the odd numbers in the way you describe.

And I can see that solution. I was just asking a simple question, if was to better to instead create a "proof" for n as any positive or negative integer that covers all cases. I guess you missed that..