Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.

Step-by-step solution

Idea: We need n(n + 1)2 > 1000, i.e. n(n + 1) > 2000. Since n(n + 1) is a bit more than n², try n near √2000 ≈ 44.7.

1. Condition: n(n + 1)2 > 1000 ⇒ n(n + 1) > 2000.1 mark
2. n = 44: 44 × 45 = 1980 < 2000, so S₄₄ = 990 (not enough).1 mark
3. n = 45: 45 × 46 = 2070 > 2000, so S₄₅ = 1035 > 1000. The smallest value is n = 45.1 mark

Check: 1 + 2 + … + 45 = 45 × 23 = 1035 ✓.

Answer to write in the exam

Common mistakes that cost marks

• Solving n(n + 1) ÷ 2 = 1000 exactly and giving a decimal. n must be a whole number, and the sum must be greater than 1000.
• Checking only n = 45 without showing that n = 44 is too small; ‘smallest’ needs both.
• Forgetting to halve: solving n(n + 1) > 1000 gives n = 32 (32 × 33 = 1056), but S₃₂ is only 528.

How this can come in the exam

Try one yourself

Find the smallest n such that 1 + 2 + … + n is greater than 2000.

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