6.EE.2 6.EE.4 6.EE.6
Act 1:
Watch the video:
Ask students what they noticed and what they wonder (are curious about). Record student responses.
Focus question(s): Have students hypothesize how the trick works. How can the performer know the sum so quickly for any number chosen? Does this work for all numbers? What would be a
Act 2:
Students work on determining how the trick works based on their hypothesis. They should be guided to show what is happening in the trick first through the use of some model that can be represented in a diagram, and then later written as an expression. Students may ask for information such as: “How were the other numbers generated after the start number was chosen?” When they ask, tell them they were the 4 consecutive numbers after the start number.
Students may ask if they can do the trick using technology: http://scratch.mit.edu/projects/20832805/
Students may also ask for materials to use (they may even ask to use similar materials from the previous task) – these can also be suggested, carefully, by the teacher.
Act 3:
Students will compare and share solution strategies.
- Share student solution paths. Start with most common strategy.
- Students should explain their thinking about the mathematics in the trick.
- Ask students to hypothesize again about whether any number would work – like fractions or decimals. Have them work to figure it out.
- Be sure to help students make connections between equivalent expressions (i.e. the rules 5n + 10 and 5(n + 2).
- Revisit any initial student questions that were not answered.
Under the Dome 
Thanks for this! I can see how the rule 5n + 10 fits the pattern. I looked at what was added to the starting number in each addend, 0+1+2+3+4, for a total of 10 added to 5 instances of the starting number). However, I’m not sure how to explain the equivalent 5(n+2) in the context of the problem. Can you help me see it? Thanks!
Hi Tiffany!
Thank you for your comment! The equivalent expression, 5(n + 2), can be seen in the sequence by looking at the number (n + 2) which is always the middle number of the five terms. This number multiplied by 5 will always give the sum of the 5 term sequence. Interestingly, it doesn’t matter what the common difference between the numbers is. In the problem here, the common difference between terms is 1, but even if the common difference is 3, the same rule applies. Ex: 4, 7, 10, 13, 16 The middle number multiplied by 5 gives the sum of 50! Isn’t math cool?
Is there a similar rule for a series with 3 or 7 or 9 terms? How about an even number of terms? This may be another blog post soon!!
I hope this helps.
Mike