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Algebraic identities · 3 marks

Consider any three consecutive square numbers. For example, 1, 4, and 9. Add the smallest and the largest squares. Thus, 1 + 9 = 10. Then subtract twice the middle square from this sum. This leads to 10 − (2 × 4) = 10 − 8 = 2. Now try the same process with another set of three consecutive square numbers. Say 9, 16, 25.

Answer: For 9, 16, 25: (9 + 25) − (2 × 16) = 34 − 32 = 2. The answer is always 2, because (n − 1)2 + (n + 1)2 − 2n2 = 2 for every number n.

Step-by-step solution

Given: Three consecutive square numbers, for example 9, 16, 25; Rule: (smallest + largest) − 2 × (middle)
To find: The result of the rule for 9, 16, 25, and whether it is always the same

Idea: Try the rule on numbers first, then write three consecutive numbers as n − 1, n, n + 1. Expanding the squares with the identities (a ± b)2 shows why the answer never changes.

  1. Try 9, 16, 25 (these are 32, 42, 52).
    Smallest + largest = 9 + 25 = 34
    Twice the middle = 2 × 16 = 32
    34 − 32 = 2½ mark
  2. Try one more set: 25, 36, 49 (52, 62, 72).
    (25 + 49) − (2 × 36) = 74 − 72 = 2 again.½ mark
  3. Why is it always 2? Any three consecutive numbers can be written as n − 1, n, n + 1. Their squares are (n − 1)2, n2, (n + 1)2.½ mark
  4. Expand the smallest and largest squares:
    (n − 1)2 = n2 − 2n + 1
    (n + 1)2 = n2 + 2n + 1
    Sum = 2n2 + 2 (the −2n and +2n cancel).1 mark
  5. Subtract twice the middle square, 2n2:
    2n2 + 2 − 2n2 = 2.
    The n has disappeared, so the answer is 2 whatever the starting number.½ mark
For 9, 16, 25 the result is (9 + 25) − 2 × 16 = 2. For any three consecutive squares the result is always 2, since (n − 1)² + (n + 1)² − 2n² = 2.

Check: Take 100, 121, 144 (10², 11², 12²): 100 + 144 = 244 and 2 × 121 = 242, so 244 − 242 = 2. ✓

Answer to write in the exam

9 + 25 − 2 × 16 = 34 − 32 = 2

25 + 49 − 2 × 36 = 74 − 72 = 2

Let the squares be (n − 1)2, n2, (n + 1)2

(n − 1)2 + (n + 1)2 − 2n2 = (n2 − 2n + 1) + (n2 + 2n + 1) − 2n2 [(a ± b)2 = a2 ± 2ab + b2]

= 2

∴ The result is always 2

Common mistakes that cost marks

  • Subtracting the middle square only once (34 − 16 = 18). The rule says subtract twice the middle square.
  • Writing (n + 1)2 = n2 + 1. The middle term 2n is missing; (a + b)2 = a2 + 2ab + b2.
  • Checking two or three examples and calling it proved. Examples only suggest a pattern; the algebra with n proves it for every case.

How this can come in the exam

MCQ (1 mark)

For the consecutive squares 64, 81, 100, the value of (64 + 100) − 2 × 81 is

  1. 0
  2. 1
  3. 2
  4. 4
Show answer

(C) 2
164 − 162 = 2. It is always 2 for three consecutive squares.

Assertion–Reason (1 mark)

Assertion (A): For any three consecutive square numbers, (smallest + largest) − 2 × middle = 2.
Reason (R): (n − 1)2 + (n + 1)2 = 2n2 + 2.

  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
Show answer

(A) Both A and R are true, and R is the correct explanation of A.
R is true by expanding both squares. Subtracting 2n2 from 2n2 + 2 leaves 2, which is exactly A.

Try one yourself

Take any three consecutive square numbers and find (largest − middle) − (middle − smallest). Is the answer always the same? Prove it.

Show answer

With (n − 1)2, n2, (n + 1)2: largest − middle = 2n + 1 and middle − smallest = 2n − 1. Difference = (2n + 1) − (2n − 1) = 2, always. (It is the same rule written another way.)

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