# Summation equation for odd numbers?

1. Feb 20, 2010

### aeonsky

$$\sum_{i=1}^{n} i$$ is the sum of all numbers between 1 and n.

I'm trying to find one for odd numbers where you need to find the sum of all odd numbers between 1 and n.

I tried 2n+1 which worked, only for first n numbers, not for numbers 1 to n.

Thanks for the help.

Last edited: Feb 20, 2010
2. Feb 20, 2010

### Staff: Mentor

Here is the sum of all the integers from 1 to an odd integer 2k + 1.
$$\sum_{i=1}^{2k + 1} i$$
To find the sum of the odd integers from 1 to 2k + 1, just subtract the sum of the even integers from the sum of all of the integers from 1 to 2k + 1.

For example, if you want the sum of the first 6 odd integers you can do this:
1 + 3 + 5 + 7 + 9 + 11 = 1 + 2 + ... + 11 - (2 + 4 + 6 + 8 + 10).

Presumably you know a formula for the sum of the first k integers. The expression in parentheses is 2(1 + 2 + 3 + 4 + 5).

3. Feb 20, 2010

### aeonsky

I don't think I am quite looking for that.

I need to find the sum of all odd integers between 1 and odd integer n.

Thus, if you need to find (3), the sum will be 4. (5), the sum will be 9. (7), the sum will be 16.

I found the formula (N+1)(N+1)/4 where n is the odd integer, but not the summation formula.

By the way, is it appropriate to have an expression for the top limit for summations?

4. Feb 20, 2010

### Mandark

If n is odd then n = 2k + 1 for some integer k. So the sum becomes $$\sum_{i=0}^{k} (2i + 1)$$. Now find k in terms of n, and hence a closed form.

5. Feb 21, 2010

### CRGreathouse

Yes.

6. Feb 21, 2010

### HallsofIvy

What Mark44 gave you will certainly help. The sum of all integers from 1 to n is, of course, n(n+1)/2. The sum of all even integers from 1 to n, assuming for the moment that n is even, n= 2k, is 2+ 4+ 6+ ...+ 2k-2+ 2k= 2(1+ 2+ 3+ ...+ (k-1)+ k) and its sum is 2[k(k+1)/2)= k(k+1). The sum of all odd numbers, from 1 to 2k+1, is the sum of all integers from 1 to 2k+1, (2k+1)(2k+2)/2= (2k+1)(k+1), minus the sum of even integers from 1 to 2k: the sum of all odd integers from 1 to 2k+1 is (2k+1)(k+1)- k(k+1)= (k+1)(2k+1-k)= (k+1)2.

Interesting result isn't it? Yes, 1+ 3= 4, 1+ 3+ 5= 9, 1+ 3+ 5+ 7= 16, etc. all squares.

Another reason why the sum of odd integers is a square:

The sum of odd numbers from 1 to 2k+1 is just an arithmetic series with "common difference" 2. An arithmetic sequence has the nice property that the average of all the numbers is just the average of the first and last numbers. Here the first and last numbers are 1 and 2k+ 1 so their average is (1+ 2k+ 1)/2= (2k+2)/2= k+ 1. There are, of course, k+1 odd numbers from 1 to 2k+1 so their sum is the average times the number of odd numbers, (k+1)(k+1)= (k+1)2.

Last edited by a moderator: Feb 21, 2010
7. Feb 21, 2010

### robert Ihnot

Examples can easily handled this problem:
1 = 1^2
1+3 = 4=2^2
1+3+5 = 9 = 3^2.
However, to prove it you can use induction.

8. Feb 29, 2012

### VeronicaS

I don't know the equation, but, the sum of odd numbers 1 to n is equal to the square of however many odd numbers your adding together. The sum of the odd numbers from 1 to 19, the first 10 odd numbers, is 100. The sum of the first 30 is 900, the first 1000 odd numbers add up to 1,000,000.

9. Feb 29, 2012

### dodo

Yet another way to get to the same place, is to imagine the sum of the first, say, 6 odd numbers, 1 + 3 + 5 + 7 + 9 + 11, as equal to the sum of the first 6 *even* numbers, 2 + 4 + 6 + 8 + 10 + 12 = 2(1 + 2 + 3 + 4 + 5 + 6)... but then subtracting one for each number you added (6, in this case).

In the general case, the sum of the first 'k' odd numbers is the sum of the first 'k' even numbers, minus 'k': that is, k(k+1) - k = k^2.

P.S. Oh gosh, the thread was 2 years old... Nevermind.

Last edited: Feb 29, 2012