Sum of all the odd numbers?

Answer 1

All right, first we need to write this in an understandable fashion. I assume you mean a sum of an odd number of odd numbers. We can write such a sum in this fashion...

SUM from i=1 to i=2n+1 of (2i+1)

Now, this can be proven to be equal to 4n2+8n+3 for some integer n. Now, doing a quick factorization, we see that this is equal to (2n+1)(2n+3), again, for integer n. Since the left side of the product (2n+1) is odd, and the right side of the product (2n+3) is also odd, we can see that (2n+1)(2n+3), being the product of two odd numbers, is also odd. Thus, the sum of any odd amount of odd numbers will also be odd.

Answer 2

The sum of two odd numbers is always an even number, e.g. 3 + 5 =8. This can be proven as follows:

Every even number can be expressed as 2m where m is a real number.

Every odd number can therefore be expressed as 2m+1 where m is a real number.

For this proof, we will consider two different odd numbers: 2m+1 and 2k+1.

2m+1 + 2k+1

=2m + 2k + 2

=2(m + k + 1)

As m+k+1 is a real number, the answer must be even. Therefore the sum of any two odd numbers is an even number.

Answer 3

All of them? If you include negative numbers then zero, as every positive number has a negative that will cancel it out. If you mean only positive numbers then it is infinity.

Answer 4

The sum of nth odd number can be calculated by S(n+1) = (n+1)(n+2)/2

Answer 5

1+3+5+7+9 = 25

Answer 6

16