Q1.
Try and find other patterns like this one. For example, you could consider 4 consecutive squares and see if you can find a pattern.
Answer
Take four consecutive squares and compare the outer pair with the inner pair. The answer is always 4.
| Four consecutive squares | (outer sum) − (inner sum) | Result |
|---|---|---|
| 1, 4, 9, 16 | (1 + 16) − (4 + 9) = 17 − 13 | 4 |
| 4, 9, 16, 25 | (4 + 25) − (9 + 16) = 29 − 25 | 4 |
| 25, 36, 49, 64 | (25 + 64) − (36 + 49) = 89 − 85 | 4 |
Now prove it. Four consecutive numbers can be written as (n − 1), n, (n + 1), (n + 2).
(n − 1)² + (n + 2)² − [ n² + (n + 1)² ]
= (n² − 2n + 1) + (n² + 4n + 4) − n² − (n² + 2n + 1)
= 2n² + 2n + 5 − 2n² − 2n − 1
= 4, whatever n is
= (n² − 2n + 1) + (n² + 4n + 4) − n² − (n² + 2n + 1)
= 2n² + 2n + 5 − 2n² − 2n − 1
= 4, whatever n is
Every term containing n cancels, so the answer cannot depend on which four squares you picked.
Why it happens: The gap between consecutive squares grows by a fixed amount. Since (k + 1)² − k² = 2k + 1, consecutive gaps are 2n − 1, 2n + 1, 2n + 3 — each 2 more than the one before. The chapter’s own pattern with three squares is exactly this constant difference of gaps, which is 2. With four squares you are comparing the last gap with the first, and they differ by two steps of 2, that is 4.
Try This: Do the same with cubes. Take four consecutive cubes and compute (fourth − third) − 2(third − second) + (second − first). You will always get 6 — and 6 = 3! is not a coincidence.