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Maths olympiad question · P5–P6★★★★ Expert

Clever sums: olympiad question 9

An expert question on clever sums for P5–P6, with a step-by-step solution, diagrams and the answer. The first few questions of a competition paper are often clever sums, and the Hua Xia Cup has them in every year from P1 to P6.

The question

It is known that 1² + 2² + 3² + … + n² = n × (n + 1) × (2n + 1) ÷ 6 (for example 1² + 2² + 3² = 3 × 4 × 7 ÷ 6 = 14).

Use it to work out the sum below the clever way.

2² + 5² + 8² + 11² + … + 17² = ?

What is the answer?

Show the hints (3)
  1. Hint 1: The bases 2, 5, 8, … go up by 3, so each is 3 × k − 1 (k = 1, 2, 3, …). Which k gives the last, 17?
  2. Hint 2: Multiply out (3 × k − 1) × (3 × k − 1): what three parts does it make? Picture a square whose side is split into 3 × k and 1.
  3. Hint 3: (3 × k − 1)² = 9 × k² − 6 × k + 1. With k from 1 to 6, find the sum of the k² and the sum of the k.

Worked solution

Have a go first, then read the solution step by step.

  1. 1.The bases go up by 3: the kth is 3 × k − 1. The last, 17 = 3 × 6 − 1, so k runs from 1 to 6: 6 terms.
  2. 2.(3 × k − 1)² = 3 × 3 × k² − 2 × 3 × 1 × k + 1 × 1 = 9 × k² − 6 × k + 1. (A big square, two strips and a small square.)
  3. 3.1² + 2² + … + 6² = 6 × 7 × 13 ÷ 6 = 91; 1 + 2 + … + 6 = 6 × 7 ÷ 2 = 21.
  4. 4.The sum = 9 × 91 − 6 × 21 + 1 × 6 = 819 − 126 + 6 = 699.
  5. ⚠️Common mistake: taking (3 × k − 1)² as 9 × k² + 1, missing the middle 6 × k and getting 825; or counting (17 − 2) ÷ 3 = 5 terms.

Answer: 699

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