Power of Two as Sum of Consecutive Integers

A sum of $k+1$ consecutive integers $n + (n+1) + (n+2) \cdots + (n+k)$ can be written as $\sum_{i=0}^{k}{n+i} = n(k+1) + \frac{(k)(k+1)}{2} = (k+1)\left(n + \frac{k}{2}\right)$ Assume there's a power of 2 that can be written as $(k+1)\left(n + \frac{k}{2}\right)$ for some natural $n$ and $k$ , then both $(k+1)$ and $\left(n + \frac{k}{2}\right)$ must be powers of 2, which means that $k$ must be odd. But that would mean $\left(n+\frac{k}{2}\right)$ is not an integer, and so not a power of 2. This is a contradiction, and so our original assumption was false. Therefore, no power of two can be written as a sum of consecutive integers and $\boxed{\text{There is no such }k}$ is the correct answer.

As with Jose's proof, we take the sum of $(k+1)$ consecutive integers (with $k \geq 1$ ), starting at $n$ , to be

$\sum_{i=0}^k (n+i) = (k+1)(n+\frac{k}{2}) = 2^d$ , for some positive integer $d$ .

Let us rewrite this: $(k+1)(\frac{2n + k}{2}) = 2^d$ , so

$(k+1)(2n+k) = 2^{d+1}$ .

Now we know that all the numbers involved are integers. But if $(k+1)$ is a factor of $2^{d+1}$ , and $k \geq 1$ , $k$ must be odd. Also, if $(2n+k)$ is a factor of $2^{d+1}$ , $k$ must be even. There's the contradiction.