Summing Consecutive Numbers

507 /www/nrich/html/content/97/05/six4/ 1 2011-07-18T13:18:57 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Watch the video below to see how numbers can be expressed as sums of consecutive numbers. <video controls="controls" height="300" src="Summing%20Consecutive%20Numbers.mp4" tabindex="0" width="400"> </video> Investigate the questions posed in the video and any other questions you come up with. Can you draw any conclusions? Can you support your conclusions with convincing arguments or proofs? If you are unable to view the video, click below to reveal an alternative version of the problem. <div class="toggle"> Charlie has been thinking about sums of consecutive numbers. Here is part of his working out: <mdo:image alt="Charlie's working" src="consec1.jpg"></mdo:image> Alison looked over Charlie's shoulder: "I wonder if we could write every number as the sum of consecutive numbers?" "Some numbers can be written in more than one way! I wonder which ones?" "$9$, $12$ and $15$ can all be written using three consecutive numbers. I wonder if all multiples of $3$ can be written in this way?" "Maybe you could write the multiples of $4$ if you used four consecutive numbers..." Choose some of the questions above, or pose some questions of your own, and try to answer them. Can you support your conclusions with convincing arguments or proofs? </div> <a href="http://nrich.maths.org/7999"> Click here for a poster of this problem</a>.</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Swaathi, from Garden International School, started by listing the numbers up to 15 and trying to represent them as sums of consecutive numbers: 2 3 = 1+2 4 5 = 2+3 6 = 1+2+3 7 = 3+4 8 9 = 4+5 = 2+3+4 10 = 1+2+3+4 11 = 5+6 12 = 3+4+5 13 = 6+7 14 = 2+3+4+5 15 = 7+8 = 4+5+6 = 1+2+3+4+5 We can't write every number as a sum of consecutive numbers - for example, 2, 4 and 8 can't be written as sums of consecutive numbers. In the above, 9 and 15 were the only numbers that I could find that could be written in more than one way. Many people spotted the pattern that all odd numbers (except 1) could be written as the sum of two consecutive numbers. For example, Matilda and Tamaris wrote: If you add two consecutive numbers together, the sum is an odd number, e.g. 1+2=3 2+3=5 3+4=7 4+5=9 5+6=11 6+7=13 and so on... Well done to pupils from Kenmont Primary School who noticed this, and explained that an Odd plus an Even is always Odd. Some spotted a similar pattern for multiples of 3. Julia and Lizzie said: If you add any 3 consecutive numbers together it will always equal a multiple of 3, e.g. 1+2+3=6 2+3+4=9 3+4+5=12 4+5+6=15 5+6+7=18 Continuing with the patterns, the Lumen Christi grade 5/6 maths extension program team sent us: We discovered that the sum of four consecutive numbers gave us the number sequence 10, 14, 18, 22, 26, 30, and so on. They were all even numbers that had an odd number as half of its total. 1+2+3+4=10 2+3+4+5=14 3+4+5+6=18... Heather from Wallington High School for Girls explained this pattern: 10 - 1+2+3+4 14 - 2+3+4+5 18 - 3+4+5+6 22 - 4+5+6+7 In all the columns, each place adds 1 each time, so in total you add 4 each time. Ruby said: Numbers which are multiples of 5, starting with 15, are sums of 5 consecutive numbers: 1+2+3+4+5=15 2+3+4+5+6=20 3+4+5+6+7=25... Fergus and Sami noticed a similar pattern: If you allow negative numbers, you can find a sum for any multiple of 7 easily. Each time you add one number either side of the sum, your sum increases by 7, e.g. 3+4=7 2+3+4+5=14 1+2+3+4+5+6=21 0+1+2+3+4+5+6+7=28 -1+0+1+2+3+4+5+6+7+8=35... Great! (There's a way to make this pattern work even without using negative numbers - can you spot it?) Why are all these patterns arising? Becky spotted a different type of pattern: We found out that powers of 2 (2, 4, 8, 16...) can never be made by adding together consecutive numbers together. Interesting! I wonder why? The Lumen Christi team give a way of constructing lots of multiples of odd numbers: We worked out that if you divide a multiple of 3 by 3, and call the answer n, then your original number is the sum of (n-1), n and (n+1). Then we discovered that the multiples of 5 can be written as 5 consecutive numbers. It's the same as the rule for 3 consecutive numbers. Take a number and divide it by 5, call it n, and then your number is the sum of (n-2), (n-1), n, (n+1) and (n+2). We then made a conjecture that since it is true for 3 and 5, it would also work for 7, 9 and any other odd number. We tested it, and it worked. For example, 63 is a multiple of 7 and 9: 7 numbers: 6+7+8+9+10+11+12=63 (63/7 = 9) 9 numbers: 3+4+5+6+7+8+9+10+11=63 (63/9 = 7) How could we take this investigation further? Arthur asked: Are there any other patterns? Can we explore the powers of two further? Is there a nice way to write certain numbers (for example, every other even number) as a sum of consecutive numbers? Ottilie suggested: Instead of adding, you could multiply the consecutive numbers, and see what patterns come up. You could also only add consecutive even numbers, or only consecutive odd numbers. These things could all have something in common, or there could be a pattern between them, or nothing at all, maybe? Magnus asked: Is the rule that the powers of two can never be made always true? Can all numbers except the powers of two be made? Great questions! By the way, Abhi sent us a nice algebraic proof that powers of 2 can never be made: 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 $2^n$. Nicely done!</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem?</h3> <div><a href="http://nrich.maths.org/public/viewer.php?obj_id=507">This problem</a> offers a simple context to begin exploration that naturally leads to some interesting conjectures about properties of numbers, and the possibility of developing some quite sophisticated algebraic arguments and proofs.</div> <h3>Possible approach</h3> <video controls="controls" height="300" src="Summing%20Consecutive%20Numbers.mp4" tabindex="0" width="400"> </video> One approach is to introduce the lesson in the way it is introduced on the video, with students suggesting numbers and then the teacher at first, and then the whole class, finding ways to express them as consecutive sums. Here is an alternative approach: "Can someone give me a set of two or more consecutive numbers?" Write a few sets on the board. "What totals do we get by adding the consecutive numbers in these sets?" Write + signs in between the lists of numbers. "These totals are all examples of numbers that can be written as the sum of consecutive numbers. Do you think all numbers can be written in this way?" "How about trying to write the numbers from $1$ to $30$ as sums of consecutive numbers?" Give students time to work in pairs on filling in the gaps from $1$ to $30$. While they are working, write the numbers from $1$ to $30$ on the board ready to collect together the sums the class have found. If students ask about negative numbers, one possible answer is: "Stick to positive numbers for now, and then perhaps investigate negative numbers later." Once most pairs have filled in most of the gaps, collect their results on the board. "Spend a minute looking at these results and then be prepared to talk about anything interesting that you notice." Give them time to think on their own at first and then share ideas with their partner, before discussion with the whole class. Next, collect together any noticings, and write them on the board in the form of questions or conjectures. If such conjectures are not forthcoming, there are some suggested lines of enquiry in the <a href="/507">problem</a>. Allow pairs time to work on the conjectures of their choosing, reminding them that they will need to provide convincing arguments to explain any of their conclusions. If appropriate, bring the class together to spend some time discussing algebraic representations of consecutive numbers ($n, n+1, n+2...$) to give students the tools to create algebraic proofs. Finally, students could create a poster, a presentation or a short report explaining one key conclusion that they came to, together with the convincing arguments they used to explain it. <h3>Key questions</h3> <div>What happens when you add together a pair of consecutive numbers? Three consecutive numbers? Four?</div> <div>What do you notice about the numbers you can't make?</div> <div>If the first of a set of consecutive numbers is $n$, how can you express the next numbers in the set, and hence the total, algebraically?</div> <div> </div> <h3>Possible extension</h3> <div>Challenge students to find an efficient way to calculate how many different ways $x$ can be expressed as a sum of consecutive numbers, for any $x$.</div> <div> </div> <div>More challenging extension - prove that it is not possible to write $2^n$ as the sum of consecutive positive whole numbers, for any $n$.</div> <div> </div> <div><a href="/2278">Pair Products</a> offers another good context for exploring properties of numbers and then using algebra to represent and explain the results.</div> <div> </div> <h3>Possible support</h3> <div> </div> <div>When collecting together the class's results for the numbers from $1$ to $30$, they can be arranged on the board in ways that will make it easier for patterns to emerge. Charlie's results from the problem could be set out like this:</div> <div> </div> <div> <table style="width: 500px;border-spacing:1px;" border="1"> <tbody> <tr> <td>9 =</td> <td>4+5</td> <td>2+3+4</td> <td> </td> <td> </td> </tr> <tr> <td>10 =</td> <td> </td> <td> </td> <td>1+2+3+4</td> <td> </td> </tr> <tr> <td>11 =</td> <td>5+6</td> <td> </td> <td> </td> <td> </td> </tr> <tr> <td>12 =</td> <td> </td> <td>3+4+5</td> <td> </td> <td> </td> </tr> <tr> <td>13 =</td> <td>6+7</td> <td> </td> <td> </td> <td> </td> </tr> <tr> <td>14 =</td> <td> </td> <td> </td> <td>2+3+4+5</td> <td> </td> </tr> <tr> <td>15 =</td> <td>7+8</td> <td>4+5+6</td> <td> </td> <td>1+2+3+4+5</td> </tr> </tbody> </table> </div> <div> </div> <div> </div></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Start by trying some simple cases; perhaps start by exploring what happens when you add two or three consecutive numbers. $1 + 2 + 3 = 6$ $2 + 3 + 4 = 9$ Can you explain why the total has gone up by $3$? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Liam noticed that: If you add an odd number to an even number it is always odd - so you can't make 2 consecutive numbers add up to an even number. Several of you, including Luke, Nikhil, Hari, Carthick, William, Daniel, Babu and Alex, noticed: You can make multiples of $3$ using $3$ consecutive numbers. Sonia from the Garden International School added: I've found out that since 1+2=3 2+3 must add up to 2 more. So if 2 consecutive numbers are added together, the totals will go up by 2. That means if 3 numbers are added together (1+2+3), the next numbers (2+3+4) will add up to 3 more. So if 3 consecutive numbers are added together, the totals will go up by 3. It means there is a pattern like 'if 4 numbers are added together, the totals will go up by 4'. Owen and Thomas said: <div>All prime numbers can only be made from $2$ consecutive sums.</div> All except the prime number $2$. Can you think why? Year 7 students at Streetfield Middle School and students from Mount Pleasant Primary School noticed that: Any number which is a power of $2$ cannot be made with consecutive sums: $2, 4, 8, 16, 32 , 64, 128...$ Rajeev from Haberdashers' Aske's Boys' School sent us <a href="/content/97/05/six4/Rajeev.docx">this</a> clear and comprehensive solution. Isabelle, Chip and Sam all had similar ideas about summing an odd number of consecutive integers. Chip explained his reasoning clearly: The consectutive integers can be found by dividing any number by one of its odd factors. For example, dividing $33$ by $3$ gives us $11$. Then, add $1$ and take $1$ on either side until there are as many numbers as the factor you used. So $10$, $11$ and $12$ are the three numbers from dividing $33$ by $3$. If you then say $10$ is $1+2+3+4$ but it is not divisible by $4$ that only works because it is actually being divided by $5$ but the zero is left out. This method works to any size number for example $91 = 13 \times7$ so it can be made from $10 + 11 + 12 + 13 + 14 + 15 + 16$. Even numbers only divide odd numbers and work correctly as $10$ divided by $2$ equals $5$ which goes to $6 + 4 = 10$ but they are not consecutive. If therefore you divide $39$, an odd, by $6$, an even, you get $6.5$. Then using some common sense you get $4$, $5$, $6$, $7$, $8$, $9$. The only way to tell if the factor will work is to see if the intended numerator is halfway between multiples of the factor. For example, with $39$ it is halfway between $36$ and $42$, both multiples of $6$ so it will be able to have six consecutive numbers add up to it. For those of you who want to investigate this problem further the following algebraic solution from Jack, Paul and Matthew might be helpful: Like the number $15$, all odd numbers can be written as the sum of two consecutive integers. For any odd number $k$, we know that $k-1$ and $k+1$ are even, also $(k-1)/2$ and$(k+1)/2$ are consecutive integers, so we can write $k=(k-1)/2+(k+1)/2$. All numbers of the form $N = pk$ where $p$ and $k$ are integers, and $k$ is odd and greater than one, can be written as the sum of consecutive integers. As before we use the integers $(k -1)/2$ and $(k +1)/2$ but this time we have to add $2p$ consecutive integers, $p$ of them less than $(k /2)$ and $p$ of them greater than $(k /2)$, so that their mean is $k /2$ and their sum is $(2 p ) \times( k /2) = pk$ For example $N = 44 = 4 \times 11$ can be written as the sum of $8$ consecutive integers, $4$ less than $5.5$ and $4$ greater than $5.5$ so that the sum is $8 \times5.5 = 44$. $44$ = ${\bf \frac{11 - 7}{2}}$+ ${\bf \frac{11 - 5}{2}}$ + ${\bf \frac{11 - 3}{2}}$ + ${\bf \frac{11 - 1}{2}}$ + ${\bf \frac{11 + 1}{2}}$ + ${\bf \frac{11 + 3}{2}}$ + ${\bf \frac{11 + 5}{2}}$+ ${\bf \frac{11 + 7}{2}}$ $= 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 $ This construction works for all numbers of the form $N = pk$ where $p$ and $k$ are integers and $k$ is odd and greater than one, but the consecutive sum may include some negative integers. If this happens then a string of terms reduces to zero giving fewer terms in the final expression. By this method, \begin{eqnarray}12 &=& 4 \times3 \\ &=& (-2) + (-1) + 0 + 1 + 2 + 3 + 4 + 5 \\ &=& 3 + 4 + 5 \end{eqnarray} giving three terms in the final expression rather than eight. If a number is a power of $2$ then it is impossible to write it as a sum of consecutive integers: Consider the sum of consecutive integers starting at $( q + 1)$ and ending at $p$ . There are $( p - q )$ terms in this sum. The mean of the terms is $( p + q + 1)/2$ so we can write: $$( q + 1) + ( q + 2) + \ldots+( p - 1) + p = ( p - q )( p + q + 1)/2$$ As $( p - q )$ and $( p + q + 1)$ cannot both be even, this means that the sum of consecutive integers must represent a number which has at least one odd factor (other than the factor $1$) so a power of $2$ cannot be written as a sum of consecutive integers. We use the fact here that if $(p - q)$ is odd then $(p + q)$ is odd and $(p + q + 1)$ is even, or if $(p - q)$ is even then $(p + q)$ is even and $(p + q + 1)$ is odd. Year 7 Set 1 at Streetfield Middle School made some interesting observations. They noticed that: Any number which is a power of $2$ cannot be made with consecutive sums. Owen and Thomas said: <div>All prime numbers only have $2$ consecutive sums.</div> All except the prime number $2$. Can you think why? Liam also made excellent observations: If you add an odd number to an even number it is always odd - so you can't make $2$ consecutive numbers add up to an even number. You can make multiples of $3$ using $3$ consecutive numbers. Titus from Garden International School tried experimenting with some different numbers including negative numbers: $30=9+10+11$ $30=6+7+8+9 $ $30=4+5+6+7+8 $ $30=(-3)+(-2)+(-1)+0+1+2+3+4+5+6+7+8 $ $30=(-5)+(-4)+(-3)+(-2)+(-1)+0+1+2+3+4+5+6+7+8+9 $ $45=22+23 $ $45=14+15+16 $ $45=7+8+9+10+11$ $45=5+6+7+8+9+10$ $45=1+2+3+4+5+6+7+8+9 $ $45=0+1+2+3+4+5+6+7+8+9$ $45=(-4)+(-3)+(-2)+(-1)+0+1+2+3+4+5+6+7+8+9+10 $ $45=(-6)+(-5)+(-4)+(-3)+(-2)+(-1)+0+1+2+3+4+5+6+7+8+9+10+11$ $60=19+20+21 $ $60=10+11+12+13+14$ $60=4+5+6+7+8+9+10+11 $ $60=(-3)+(-2)+(-1)+0+1+2+3+4+5+6+7+8+9+10+11$ $60=(-9)+(-8)+(-7)+(-6)+(-5)+(-4)+(-3)+(-2)+(-1)+0+1+2+3+4+5+6+7+8+9+10+11+12+13+14 $ Sonia and Yun Seok, also at Garden International School, reached some interesting conclusions: I've found out that all the odd numbers except number $1$ had at least one consecutive integer. I also think that numbers from $17$ and above have at least one consecutive integer too. I've also found out that $1+2=3$ so $2+3$ must add up $2$ more. So if $2$ numbers are added together, the next numbers are added up $2$ more. That means if $3$ numbers are added together $(1+2+3)$, the next numbers add up to $3$ more. It means there is a pattern like 'if $4$ numbers are added together, the next numbers are added up $4$ more' All odd numbers can be written in two consecutive numbers added up together. All numbers in any of the odd timetables can be represented in the number of the timetables they are in. So, all of the even timetables will not work. For example, $20$ is in both the $4$ times table and the $5$ times table. $20$ is the lowest common multiple (LCM) of $4$ and $5$ $20=2+3+4+5+6$ This works because when you divide $20$ by $5$ you get the middle number $4$ of the consecutive numbers. However, when you divide $20$ by $4$ you get $5$ but this cannot be in the middle of the consecutive sum so you have either: $4+5+6+7=22$ or $3+4+5+6=18$ which does not work because there cannot be a middle number for a number of even numbers. Chip from King's Ely sent in a very well thought out solution: The consectutive integers can be found by dividing any number by one of its odd factors. For example dividing $33$ by $3$ to give $11$. Then adding $1$ or taking one on either side until there are as many numbers as the factor you used. For example, $10$, $11$ and $12$ are the three numbers from dividing $33$ by $3$. If you then say $10$ is $1+2+3+4$ but it is not divisible by $4$ that only works because it is actually being divided by $5$ but the zero is left out. This method works to any size number for example $91 = 13 \times7$ so it can be made from $10 + 11 + 12 + 13 + 14 + 15 + 16$. Even numbers only divide odd numbers and work correctly as $10$ divided by $2$ equals $5$ which goes to $6 + 4 = 10$ but they are not consecutive. If therefore you divide $39$, an odd, by $6$, an even, you get $6.5$. Then using some common sense you get $4$, $5$, $6$, $7$, $8$, $9$. The only way to tell if the factor will work is to see if the intended numerator is halfway between multiples of the factor. For example, with $39$ it is halfway between $36$ and $42$, both multiples of $6$ so it will be able to have six consecutive numbers add up to it. For those of you who want to investigate this problem further the following algebraic solution from Jack, Paul and Matthew might be helpful: Like the number $15$, all odd numbers can be written as the sum of two consecutive integers. For any odd number $k$, we know that $k-1$ and $k+1$ are even, also $(k-1)/2$ and$(k+1)/2$ are consecutive integers, so we can write $k=(k-1)/2+(k+1)/2$. All numbers of the form $N = pk$ where $p$ and $k$ are integers, and $k$ is odd and greater than one, can be written as the sum of consecutive integers. As before we use the integers $(k -1)/2$ and $(k +1)/2$ but this time we have to add $2p$ consecutive integers, $p$ of them less than $(k /2)$ and $p$ of them greater than $(k /2)$, so that their mean is $k /2$ and their sum is $(2 p ) \times( k /2) = pk$ For example $N = 44 = 4 \times 11$ can be written as the sum of $8$ consecutive integers, $4$ less than $5.5$ and $4$ greater than $5.5$ so that the sum is $8 \times5.5 = 44$. \begin{eqnarray}44 &=& (11 - 7)/2 + (11 - 5)/2 + (11 - 3)/2 + (11 - 1)/2 + (11 + 1)/2 + (11 + 3)/2 + (11 + 5)/2 + (11 + 7)/2 \\ &=& 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 \end{eqnarray} This construction works for all numbers of the form $N = pk$ where $p$ and $k$ are integers and $k$ is odd and greater than one, but the consecutive sum may include some negative integers. If this happens then a string of terms reduces to zero giving fewer terms in the final expression. By this method, \begin{eqnarray}12 &=& 4 \times3 \\ &=& (-2) + (-1) + 0 + 1 + 2 + 3 + 4 + 5 \\ &=& 3 + 4 + 5 \end{eqnarray} giving three terms in the final expression rather than eight. If a number is a power of $2$ then it is impossible to write it as a sum of consecutive integers: Consider the sum of consecutive integers starting at $( q + 1)$ and ending at $p$ . There are $( p - q )$ terms in this sum. The mean of the terms is $( p + q + 1)/2$ so we can write: $$( q + 1) + ( q + 2) + \ldots+( p - 1) + p = ( p - q )( p + q + 1)/2$$ As $( p - q )$ and $( p + q + 1)$ cannot both be even, this means that the sum of consecutive integers must represent a number which has at least one odd factor (other than the factor $1$) so a power of $2$ cannot be written as a sum of consecutive integers. We use the fact here that if $(p - q)$ is odd then $(p + q)$ is odd and $(p + q + 1)$ is even, or if $(p - q)$ is even then $(p + q)$ is even and $(p + q + 1)$ is odd. The solution below may be helpful to Melanie of West Flegg Middle School, who sent in a very well reasoned, part solution to the problem. Jack, Paul and Matthew also sent in a correct solution, and again, the reasoning below may help to explain why it works. Like the number $15$, all odd numbers can be written as the sum of two consecutive integers. For any odd number $k$, we know that $k-1$ and $k+1$ are even, also $(k-1)/2$ and$(k+1)/2$ are consecutive integers, so we can write $k=(k-1)/2+(k+1)/2$. All numbers of the form $N = pk$ where $p$ and $k$ are integers, and $k$ is odd and greater than one, can be written as the sum of consecutive integers. As before we use the integers $(k -1)/2$ and $(k +1)/2$ but this time we have to add $2p$ consecutive integers, $p$ of them less than $(k /2)$ and $p$ of them greater than $(k /2)$, so that their mean is $k /2$ and their sum is $(2 p ) \times( k /2) = pk$ For example $N = 44 = 4 \times 11$ can be written as the sum of $8$ consecutive integers, $4$ less than $5.5$ and $4$ greater than $5.5$ so that the sum is $8 \times5.5 = 44$. \begin{eqnarray}44 &=& (11 - 7)/2 + (11 - 5)/2 + (11 - 3)/2 + (11 - 1)/2 + (11 + 1)/2 + (11 + 3)/2 + (11 + 5)/2 + (11 + 7)/2 \\ &=& 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 \end{eqnarray} This construction works for all numbers of the form $N = pk$ where $p$ and $k$ are integers and $k$ is odd and greater than one, but the consecutive sum may include some negative integers. If this happens then a string of terms reduces to zero giving fewer terms in the final expression. By this method, \begin{eqnarray}12 &=& 4 \times3 \\ &=& (-2) + (-1) + 0 + 1 + 2 + 3 + 4 + 5 \\ &=& 3 + 4 + 5 \end{eqnarray} giving three terms in the final expression rather than eight. If a number is a power of $2$ then it is impossible to write it as a sum of consecutive integers. Consider the sum of consecutive integers starting at $( q + 1)$ and ending at $p$ . There are $( p - q )$ terms in this sum. The mean of the terms is $( p + q + 1)/2$ so we can write: $$( q + 1) + ( q + 2) + \ldots+( p - 1) + p = ( p - q )( p + q + 1)/2$$ As $( p - q )$ and $( p + q + 1)$ cannot both be even, this means that the sum of consecutive integers must represent a number which has at least one odd factor (other than the factor $1$) so a power of $2$ cannot be written as a sum of consecutive integers. We use the fact here that if $(p - q)$ is odd then $(p + q)$ is odd and $(p + q + 1)$ is even, or if $(p - q)$ is even then $(p + q)$ is even and $(p + q + 1)$ is odd. </mdoxml> 2 3 0 0 1 0 0 Summing Consecutive Numbers Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way? Admin Workshop Admin Learning through exploration Numbers and the Number System Properties of numbers Using, Applying and Reasoning about Mathematics Working systematically Algebra Creating expressions/formulae Using, Applying and Reasoning about Mathematics Generalising Information and Communications Technology Video Information and Communications Technology smartphone Collections Secondary Number Play Secondary Mapping Document Patterns and sequences LS ajk44 solution needs editing Secondary Mapping Document DisplayCabinet Secondary processes PM - Reasoning, Justifying. Convincing. Proof. 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