My Math Forum Sum of Odd Integers

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 July 26th, 2012, 11:45 AM #1 Newbie   Joined: Jun 2012 Posts: 13 Thanks: 0 Sum of Odd Integers Let $c_1,c_2,...,c_3$ be odd integers. Prove that $\Sigma c_i$, where the sum is from i=1 to i=k, is odd iff k is odd. Thanks for help!
 July 26th, 2012, 11:49 AM #2 Global Moderator   Joined: May 2007 Posts: 6,528 Thanks: 589 Re: Sum of Odd Integers Easy by induction: Even number + odd number is odd, odd number + odd number is even. k=1, sum is odd, k=2, sum is even, and odd - even alternate as you add each term.
 July 26th, 2012, 12:00 PM #3 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: Sum of Odd Integers If k is even, then you can split the numbers into pairs, and each pair of odd numbers adds to an even number. Since all of these numbers are divisible by 2, the sum of all the numbers is also divisible by 2, and is therefore even. If k is odd, then you can split the numbers into pairs as before, except you will have one left over. The sum of all the numbers put into pairs is even, and then adding the leftover number, which is odd, gives you an odd total.
July 27th, 2012, 04:34 AM   #4
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Re: Sum of Odd Integers

Quote:
 If $c_1,c_2,c_3,...$ are odd integers, Prove that $\Phi= \sum_{i=1}^{k} c_i$ is odd iff k is odd.
Consider the sum:

$S= \sum_{i=1}^{k} \,\, = \underbrace{1\,+\,1+\,1+\,1+\,1\,+\,1\,+\,.\,.\,.} _{\text{k times}} = k \cdot 1 = k$

So, the sum S is odd if and only if k is odd.

Every term of $\Phi$ can be represented as $c_n= 2 a_n +1$

So, rewrite the sum using the above representation :

$\Phi= \sum_{i=1}^{k} c_i = \sum_{i=1}^{k} $2 a_i + 1$ = 2 \, \Sigma a_i \,+\, S$

so, $\Phi$ is odd if and only if S is odd and S is odd if and only if k is odd implies

$\Phi$ is odd iff k is odd

$\text{QED}$

 July 27th, 2012, 01:28 PM #5 Newbie   Joined: Jun 2012 Posts: 13 Thanks: 0 Re: Sum of Odd Integers Thanks for the responses, very helpful.
 July 27th, 2012, 02:29 PM #6 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Sum of Odd Integers The way I would is to note that we can write any odd number in the form "2k+1" for some integer k. That means we can write the $c_i= 2k_i+ 1$ so that the sum is $\sum_{i=0}^n 2k_i+ 1= 2\sum_{i=0}^n k_i+ n$. The first sum is even because of the "2". If n is even, this is the sum of two even numbers so even. If n is odd, this is the sum of an even and an odd number so is odd.
July 29th, 2012, 07:05 AM   #7
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Re: Sum of Odd Integers

Quote:
 Originally Posted by HallsofIvy The way I would is to note that we can write any odd number in the form "2k+1" for some integer k. That means we can write the $c_i= 2k_i+ 1$ so that the sum is $\sum_{i=0}^n 2k_i+ 1= 2\sum_{i=0}^n k_i+ n$. The first sum is even because of the "2". If n is even, this is the sum of two even numbers so even. If n is odd, this is the sum of an even and an odd number so is odd.
Isn't your method exactly same as I gave before?

 July 29th, 2012, 07:38 AM #8 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Sum of Odd Integers In some sense all methods are equivalent. But it's a somewhat different expression. The hope, presumably, is that the OP would find one of these expressions more understandable than the others.
 July 31st, 2012, 12:09 AM #9 Newbie   Joined: Jul 2012 Posts: 2 Thanks: 0 Re: Sum of Odd Integers It is difficult for me to understand it.
July 31st, 2012, 03:39 AM   #10
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Re: Sum of Odd Integers

Quote:
 Originally Posted by jicheng It is difficult for me to understand it.
What's difficult for you to understand? The solution I did?

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### two alternate odd integers

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