Four Minutes, Five Series: Reading a Student's Number-Sense Strategy
Ms. Ibarra is reviewing a fifth grader's number-series sheet. Devon finished all five items in under four minutes and wrote an answer for every one without crossing anything out. Items and his answers: (A) 3, 7, 15, 31, 63, ___ answered 127. (B) 2, 5, 10, 17, 26, ___ answered 35. (C) 100, 50, 25, 12.5, ___ answered 6. (D) 1, 1, 2, 3, 5, 8, 13, ___ answered 21. (E) 4, 6, 10, 18, 34, ___ answered 50. On the last unit Devon scored 9/10 on multiplication facts and 8/10 on two-step word problems, so Ms. Ibarra does not believe the errors come from weak arithmetic. She has ten minutes before the next session and wants one or two follow-up series that will expose exactly which strategy he is using.
1. Give the correct answer for each item and state the rule in a single sentence.
Read the worked answer
A - 127. Each term is double the previous term plus 1: 3×2+1=7, 7×2+1=15, 15×2+1=31, 31×2+1=63, so 63×2+1=127. The first differences (4, 8, 16, 32) double, which reaches the same answer - a useful self-check the item builds in.
B - 37. The first differences are 3, 5, 7, 9 - consecutive odd numbers. The next difference is 11, and 26+11=37.
C - 6.25. Each term is half the one before: 100→50→25→12.5→6.25.
D - 21. This is the Fibonacci pattern: each term is the sum of the two before it, and 8+13=21.
E - 66. Each term is double the previous term minus 2: 4×2−2=6, 6×2−2=10, 10×2−2=18, 18×2−2=34, so 34×2−2=66. The differences (2, 4, 8, 16) double, so the next difference is 32 and 34+32=66.
2. Diagnose Devon's strategy from B and E, then explain what A, C, and D tell you about the limits of that diagnosis.
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In B, Devon wrote 35 - that is 26 + 9, repeating the most recent difference instead of continuing the odd-number sequence. In E, he wrote 50 - that is 34 + 16, again repeating the last difference. Same move twice: compute the last step, then assume every step looks like it. Call it the last-difference habit.
The other three items bound that diagnosis. In A the last difference is 32, and 63 + 32 = 95, not 127; he wrote 127, so he did not apply the habit where a doubling pattern was visible. In D he added the two preceding terms correctly, which is an additive rule he had to invent, not a pattern he could copy. So the habit is conditional: when a multiplicative rule is obvious he uses it, and when it is not he falls back on 'add the last difference again.'
C is a different kind of slip. He halved correctly four times but recorded 6 instead of 6.25 - a decimal truncation, not a pattern error. Counting it as evidence for the habit would overstate the case. The arithmetic scores are consistent with this reading: nothing here requires a fact he lacks.
3. Design one follow-up series that separates a correct pattern-continuer from a last-difference reasoner, and show what each would write.
Read the worked answer
Use: 5, 6, 9, 14, 21, ___
Differences are 1, 3, 5, 7 - consecutive odd numbers again, but starting from a different place than item B, so a memorized answer will not transfer. The next difference is 9, so the answer is 30.
A last-difference reasoner computes 21 + 7 = 28. The two candidate answers are close, which is the point: a lucky guess is unlikely to land on 30 twice in a row, and the gap between 28 and 30 tells you the student evaluated a single difference rather than the sequence of differences.
A second probe if time allows: 2, 3, 5, 8, 12, ___ (differences 1, 2, 3, 4; next is 5, answer 17; a last-difference reasoner writes 16). Administer both, ask Devon to say his rule aloud before writing, and listen for whether he describes one step or a sequence. If he can state the rule but still writes the last-difference answer, the problem is execution under speed, not understanding, and the fix is a checking routine rather than more practice items.