# Impossible Sums - Proof 3

##### Age 14 to 18Challenge Level

The problem Impossible Sums asks you to think about which numbers can be written as a sum of consecutive numbers, and which numbers cannot.

Below is a proof that the sum of consecutive numbers cannot be a power of $2$. This proof uses some facts about triangular numbers.

The $n^{\text{th}}$ triangular number is the sum of the first $n$ consecutive numbers, i.e. the $n^{\text{th}}$ triangular number is $T_n=1+2+3+\cdots + (n-1) + n$.

We can write the sum of consecutive numbers as the difference between two triangular numbers, for example:

$\phantom{=(}6+7+8+9+10+11+12$
$= (1+2+3+4+5+6+7+8+9+10+11+12)-(1+2+3+4+5)$
$=T_{12}-T_5$

Before you start thinking about this proof, you may like to look at the problem Picturing Triangular Numbers, which helps to explain why the $n^{\text{th}}$ triangular number is given by the formula $\frac 1 2 n(n+1)$.

The statements below have been scrambled up. Can you rearrange them into their original order?  The first, second and last statements have been fixed in place.