TITLE: Show that   $2^n$ is not a sum of consecutive positive integers
QUESTION [5 upvotes]: Possible Duplicate:
A proof that powers of two cannot be expressed as the sum of multiple consecutive positive integers that uses binary representations? 

Suppose we have an arithmetic sequence $a_n= a_1 + (n-1)d$ such that $a_i \in \mathbb{N}$. If we let $d=1$, then we have consecutive natural numbers. Now, we are guaranteed that $$\sum_{i=1}^{k}{a_i}= \lambda, \lambda  \in \mathbb{N}$$
Now, all elements of $\mathbb{N}$ are expressible into some $\sum_{i=1}^{k}{a_i}$ but it is not unique. Say $\lambda = 15 = 1+2+3+4+5= 7+8.$ But if $\lambda = 2^n $ $\forall$ $ n \in \mathbb{N}^* $, there is no $\sum_{i=1}^{k}{a_i}$ possible where $d=1$. 
How can we show that all natural numbers of the form $2^n $ $\forall$ $ n \in \mathbb{N}^* $ cannot be written as sums of consecutive natural numbers?

REPLY [2 votes]: The claim as stated is false. If we stipulate that the sequence has length greater than $1$ and all the terms are positive, then it is true.
Consider the identity
$$
\sum_{k=m}^n k=\frac12(n+m)(n-m+1)\tag{1}
$$
holds because the sum consists of $n-m+1$ numbers whose average is $\frac12(n+m)$.
Note that $(n+m)-(n-m+1)=2m-1$ is odd. Thus, one or the other of $n+m$ or $n-m+1$ must be odd.
Suppose the sum in $(1)$ is $2^j$. The only odd factor of $2^j$ is $1$, so either $n-m+1=1$ or $n+m=1$.
Case $\mathbf{1}$: $\mathbf{n-m+1=1}$
If $n-m+1=1$, then n=m and the sum is a singleton, $n$. If $n=2^j$ this contradicts the claim unless we stipulate a sequence has a length greater than $1$.
If the sequence has more than one term, then we cannot have Case $1$.
Case $\mathbf{2}$: $\mathbf{n+m=1}$
If $n+m=1$, then one of the terms must be less than $1$. Furthermore, the sum is then $\frac12(n-m+1)=n$ and again we contradict the claim if $n=2^j$. For this case we can have
$$
\begin{align}
1&=0+1\\
2&=-1+0+1+2\\
4&=-3+-2+-1+0+1+2+3+4
\end{align}
$$
as counterexamples.
If the sequence is all positive, then we cannot have Case $2$.
The problem as stated allows Case $1$, but not Case $2$. However, if the sequence has more than one term and is all positive, then the claim is true.