Try and find other patterns like this one. For example, you could consider 4 consecutive squares and see if you can find a pattern.
Step-by-step solution
To find: A rule that always gives the same answer, and a proof of it
Idea: “This one” is the pattern for three consecutive squares: (smallest + largest) − 2 × (middle) is always 2 (e.g. 1 + 9 − 2 × 4 = 2). For four squares, pair the outside squares against the inside squares in the same way. Then write the four numbers as n, n + 1, n + 2, n + 3 and expand with (a + b)2 = a2 + 2ab + b2 to see why the answer never changes.
- Try numbers first. Take 1, 4, 9, 16:
(smallest + largest) − (two middle ones) = (1 + 16) − (4 + 9) = 17 − 13 = 4.½ mark - Another set: 16, 25, 36, 49:
(16 + 49) − (25 + 36) = 65 − 61 = 4 again.
And 100, 121, 144, 169: 269 − 265 = 4.½ mark - Prove it. Let the four consecutive numbers be n, n + 1, n + 2, n + 3. Their squares are
n2, n2 + 2n + 1, n2 + 4n + 4, n2 + 6n + 9.1 mark - Outside pair: n2 + (n2 + 6n + 9) = 2n2 + 6n + 9
Inside pair: (n2 + 2n + 1) + (n2 + 4n + 4) = 2n2 + 6n + 5½ mark - Difference = (2n2 + 6n + 9) − (2n2 + 6n + 5) = 4.
All the n terms cancel, so the answer is 4 for every set of four consecutive squares.½ mark
Check: Take 49, 64, 81, 100 (7² to 10²): 49 + 100 = 149 and 64 + 81 = 145; 149 − 145 = 4. ✓
Answer to write in the exam
(1 + 16) − (4 + 9) = 4; (16 + 49) − (25 + 36) = 4
Let the squares be n2, (n + 1)2, (n + 2)2, (n + 3)2
n2 + (n + 3)2 − (n + 1)2 − (n + 2)2 = (2n2 + 6n + 9) − (2n2 + 6n + 5)
= 4
∴ For any four consecutive squares, (smallest + largest) − (sum of the two middle squares) = 4
Common mistakes that cost marks
- Writing (n + 3)2 = n2 + 9. The middle term 2 × n × 3 = 6n is missing.
- Taking squares that are not consecutive (for example 1, 9, 25, 49). The pattern only works for squares of consecutive numbers.
- Stopping after checking one example. One example only suggests a rule; the algebra with n proves it.
How this can come in the exam
For the squares 36, 49, 64, 81, the value of (36 + 81) − (49 + 64) is
- 2
- 4
- 6
- 8
Show answer
(B) 4
117 − 113 = 4. It is always 4 for four consecutive squares.
Show that (n + 3)2 − (n + 2)2 − (n + 1)2 + n2 = 4 for every number n.
Show answer
Expand: (n2 + 6n + 9) − (n2 + 4n + 4) − (n2 + 2n + 1) + n2 (1 mark). n2 terms: 1 − 1 − 1 + 1 = 0; n terms: 6 − 4 − 2 = 0; numbers: 9 − 4 − 1 = 4. So the value is 4 (1 mark).Try one yourself
Take any two consecutive square numbers, for example 25 and 36. Show that their difference is always an odd number, and find a formula for it.
Show answer
(n + 1)2 − n2 = n2 + 2n + 1 − n2 = 2n + 1, which is always odd. For 25 and 36: n = 5, so 2(5) + 1 = 11 = 36 − 25 ✓.
More questions like this
- Think of numbers a and b where a and b do not represent lengths of line segments. What if a and b are negative numbers? Let us check for some negative numbers and see if this equation still works.
- What is the difference between an equation and an identity?
- By now you must have observed that (a + b)2 ≠ a2 + b2. But can you find out which one of them will be greater? For example, if a = 10 and b = 2, what are the values of (a + b)2 and a2 + b2?
- 1. What can you say about a and b if (a + b)2 < a2 + b2?
2. What can you say about a and b if (a + b)2 > a2 + b2?
3. When will (a + b)2 be equal to a2 + b2?
Did you observe that (a + b)2 and a2 + b2 are both positive? What term will decide which is larger? Use the expansion of (a + b)2 to decide. - Let us try to expand (5x + 2y)2.