Can you find the sum of the first ten natural numbers without actually adding all of them?

Step-by-step solution

Idea: Pair the numbers from the two ends: 1 + 10, 2 + 9, 3 + 8, … all equal 11. Writing the sum twice (once reversed) gives 10 such pairs.

1. S = 1 + 2 + 3 + … + 10 and also S = 10 + 9 + 8 + … + 1.½ mark
2. Add the two lines column by column: each column is 11, and there are 10 columns. So 2S = 10 × 11 = 110.1 mark
3. S = 110 ÷ 2 = 55.½ mark

Check: Group as five pairs (1 + 10) + (2 + 9) + (3 + 8) + (4 + 7) + (5 + 6) = 5 × 11 = 55 ✓.

Answer to write in the exam

Common mistakes that cost marks

• Forgetting to halve: 10 × 11 = 110 is twice the sum.
• Counting 11 pairs instead of 10. There are as many columns as numbers.
• Multiplying 10 × 10 ÷ 2 = 50. Each pair adds to 11, not 10.

How this can come in the exam

Try one yourself

Find 1 + 2 + 3 + … + 13 by the pairing method.

More questions like this

• Can the same approach be used to find the sum of 1 + 2 + 3 + … + 100?
• If we use the notation S_n to represent the sum of the first n natural numbers, then S_n = n(n + 1)2. Can you use this formula to find S₂₀, S₅₀ or S₁₀₀₀?
• Let us revisit the sequence t_n of triangular numbers 1, 3, 6, 10, 15, … shown in the figure. Note that the nth term of this sequence is the sum of the first n natural numbers. Thus t_n = n(n + 1)2. Can you use this to find the 10th, 17th and 80th triangular numbers?
• Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ….
• Which term of the AP : 21, 18, 15, … is −81? Also, is 0 a term of this AP? Give reasons for your answer.

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