How to prove all positive odd numbers in $3x + 1$ will result in even numbers? I was wondering, is there is a generic way to prove that for all positive odd numbers the expression $3x + 1$  will result in even number? I was checking in a brute force way, by writing a program for odd numbers ranging from 1 till Max_long and check if there are numbers which could still result in odd number.
Is there a way to mathematically prove the same ? 
 A: Do note that $3x + 1$ is an expression, not an equation. Equations have equality somewhere.

As far as proving the statement, notice that the product of two odd numbers is odd, so $3x$ is odd. Then the sum of odd numbers is even, so $3x + 1$ is even. It is a good instructional experience to prove these claims from the definition: Start with $a = 2n + 1$ and $b = 2m + 1$ as your odd numbers, and show that the product $ab$ is of the form $2(...) + 1$ and the sum is of the form $2(...)$.
A: What you're trying to prove is that if $x$ is odd then $3x+1$ is even. We know that odd numbers are of the form $2k+1, k \in \mathbb{Z}$, and even numbers are of the form $2k, k \in \mathbb{Z}$. So plug in the odd number and see if it is of the form of an even number.
Substitute
$$x=2k+1,k \in \mathbb{Z}$$
into
$$3(2k+1)+1=6k+3+1=6k+4=2(3k+2)$$
Hence the expression results in even numbers if $x$ is odd.
A: For $3x+1$ to be odd $x$ has to be even starting from $0$ .let us replace $x$ by $r$ so $\sum 3r+1=\frac{3(n)(n+1)}{2}+n=\frac{(n)(3n+5)}{2}$. So summation is always even  independent of $n$ i hope you know the reason.
