Consecutive sums/Consecutive sums activity: Difference between revisions

Questions that may be asked:

“If you add up consecutive integers what answers can you make?”

Having established the ‘rules’, we then look at formulating a question.  

I often find it useful for pupils to be able to try a few sets of numbers for themselves (what the pupils describe as “random numbers”). Having tried a few out we can then discuss how to work more systematically.

Some pupils want to make a list and try to find ways to make each number, starting from 1 …

Others have the idea of narrowing the task to adding up two consecutive numbers. This one is often more fruitful and the idea of simplifying a task to gain more insights is an important one.

'''Extension/Further work'''

Two consecutive numbers always give an odd answer. This can be explaining using ‘odd + even is odd’, or using some algebra (''n'' + ''n'' + 1 <nowiki>=</nowiki> 2''n'' + 1. ''n'' is an integer, so 2''n'' is even, which makes 2''n'' + 1 odd). Turning this round, if I tell you an odd number you can tell me the two numbers to add by halving it and taking the two integers either side of it. Eg: 27 ÷ 2 <nowiki>=</nowiki> 13.5, so we use 13+14.

Three consecutive numbers throw up some interesting things. Some pupils miss out some of the possibilities. They start out with 1+2+3 and then try 3+4+5 or 4+5+6, missing out 2+3+4. Some, when they include every possibility, will remark that the answers alternate between being even and odd. Of more use is that the answers are always multiples of 3. Every multiple of three (above 3) can be made in this way. If you divide a multiple of 3 by 3 you can then take the numbers either side of the answer and the three numbers you have add up to the original number. Eg: 33÷3 <nowiki>=</nowiki> 11, so we use 10+11+12. This can be explained in lots of ways, including referring to working out the mean of a set of numbers. Algebra can also help: the three numbers can either be treated as ''m'', (''m''+1) and (''m''+2), which have a total of 3''m''+3, or as (''n''-1), ''n'' and (''n''+1), which have a total of 3''n''. An alternative, geometric way of exploring and explaining this is shown later.

Four consecutive numbers have an even total, but are not multiple of four, whereas five consecutive numbers give a multiple of five. This can lead to hypotheses about odd and even numbers.

'''Further work/other ways of teaching this topic'''

The original NRICH version of this problem stated:

Many numbers can be expressed as the sum of two or more consecutive integers.

What can you say about numbers which can be expressed in this way?  

The most recent version is now called ''Summing Consecutive Numbers'' and is at [http://nrich.maths.org/507 http://nrich.maths.org/507]

[[File:Steps1.png]]

This means that every consecutive sum answer can be considered to be a triangular number plus a rectangle that has the same height as the triangle.

Depending on how the task is presented, some pupils think they have found a way of ‘breaking’ the problem. If you are allowed to include negative numbers you can make every single answer by summing consecutives. For example, 4 can be made as: 4 <nowiki>=</nowiki> -3 + -2 + -1 + 0 + 1 + 2 + 3 + 4. Either this can be excluded explicitly in the wording of the question (the Steps version clearly does not permit this!) or it can be left for the pupils to find, and the statement of the question altered if/when they do.

'''Assessment'''

After pupils have completed their own work I sometimes use the NRICH solutions with them.

One way to do this is to give them copies of the full solutions and to ask them to read them through. Then they need to position their own work with the solutions. Is it the same as someone else’s work? Is it more complex than some? What could they have done next to make their work even better? Are there any parts of the solutions they don’t understand? I print these off, often on A3 paper, and give them to pupils to work on in pairs. The older solutions from a previous version of the NRICH problem are available here as a Word document. The most recent solutions are on the NRICH site at: [http://nrich.maths.org/507/solution http://nrich.maths.org/507/solution]

An alternative is to cut out the different parts of the NRICH solution and to provide them on individual cards. The pupils need to assemble them in order of complexity themselves.

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