How can I show that $n=k+(k+1)+\cdots+(k+m)$

Question

How can i show that every $n \in \mathbb{N}$ can be written as summing up to $n$ togheter with some succeeding consecutive numbers?

In other way:

$$\forall n \in \mathbb{N} \ \exists k,m \in \mathbb{N} \text{ such that } n=k+(k+1)+\cdots+(k+m)$$

Thanks

Answer 1

This is not true for $n=2^\alpha$. Since if $$ n = k+(k+1)+...+(k+m), $$ then $$ 2^{\alpha+1} = 2k(m+1)+m(m+1)=(2k+m)(m+1). $$ As $(2k+m)+(m+1) = 2k+2m+1$ is an odd number, one of them must be one which is not possible since $2k+m>1$ and $m+1 >1$.

If $n$ be an odd number, it is very straightforward. So let $n=2^\alpha\beta$, where $\beta>1$ is an odd number. If $2^\alpha>\beta$, then set $m=\beta-1$ and $k=2^\alpha+\frac{1}{2}(\beta-1)$. If $2^\alpha<\beta$, set $m=2^\alpha-1$ and $k = \frac{1}{2}(\beta-2^\alpha+1)$.

Thus, the claim is true for all natural numbers which are not a power of $2$.