# Consecutive Sums Poster

## Problem

This poster is based on the problem Summing Consecutive Numbers.

The poster is available as a PDF, or the image below can be clicked on to enlarge it.

## Student Solutions

All numbers except for powers of $2$ are sums of consecutive numbers.

**Why?**

Every odd number is the sum of the two consecutive numbers closest to half of the number:

9 = 4 + 5, 47 = 23 + 24, 101 = 50 + 51, 583 = 291 + 292

Multiples of 3 can be written as the sum of 3 consecutive numbers:

15 = 4 + 5 + 6, 18 = 5 + 6 + 7, 21 = 6 + 7 + 8, 3$n$ = ($n-$1) + $n$ + ($n$+1)

This is the same for multiples of any odd number:

5$n$ = ($n-$2) + ($n-$1) + $n$ + ($n$+1) + ($n$ + 2),

7$n$ = ($n-$3) + ($n-$2) + ($n-$1) + $n$ + ($n$+1) + ($n$ + 2) + ($n$ + 3)

$(2r+1)n$ = ($n-r$) + ... + ($n-$1) + $n$ + ($n$+1) + ... + ($n$ + $r$)

Sometimes, the first terms in this addition may be negative, resulting in a shorter sum. For example, 27 = 3$\times$9

Therefore 27 = $-$1 + 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7, which we can write as 2 + 3 + 4 + 5 + 6 + 7

For even numbers which are not multiples of 4, there is another sum of consecutive numbers.

For example, 26 = 2$\times$13 = 2$\times$(6 + 7) = 6 + 6 + 7 + 7 = 5 + 6 + 7 + 8

Algebraically, even number = 2$\times m$ where $m$ is odd,

the number = 2$\times$($r$ + ($r$ + 1)) = ($r-$1) + $r$ + ($r$ + 1) + ($r$ + 2)

In fact, this works even for factors greater than 2. If the even number is 3$\times m$, we get

($r-$2) + ($r-$1) + $r$ + ($r$ + 1) + ($r$ + 2) + ($r$ + 3), and so on.

Powers of 2

For even numbers with no odd factors, none of the methods above work.

__Case 1: can we make $2^n$ from an odd number of consecutive numbers?__

An odd number of consecutive numbers has a whole number as an average. This average is always the middle number. So, that means that the sum of the numbers will be:

Sum = average $\times$ number of consecutive numbers.

= whole number $\times$ odd number

This means the sum has an odd number as a factor. But $2^n$ cannot have an odd number as a factor. This proves that an odd number of consecutive numbers cannot add to make $2^n$.__Case 2: can we make $2^n$ from an even number of consecutive numbers?__

An even number of consecutive numbers will not have a whole number as an average. The average will be the average of the two middle numbers. So:

Sum = (sum of two middle numbers) $\times \frac{1}{2} \times$ number of consecutive numbers

= (sum of two consecutive numbers) $\times$ ($\frac{1}{2}\times$ Even number)

= (sum of two consecutive numbers) $\times$ whole number

But if you add two consecutive numbers, the answer is always an odd number. So a sum like this must have an odd number as a factor again - but $2^n$ doesn't. This proves that an even number of consecutive numbers cannot add to make $2^n$.