Can the same approach be used to find the sum of 1 + 2 + 3 + … + 100?

Step-by-step solution

Idea: Reverse the sum and add it to itself: every column adds to (first + last) = 1 + 100 = 101.

1. S = 1 + 2 + … + 99 + 100 and S = 100 + 99 + … + 2 + 1.½ mark
2. Adding column by column: 1 + 100 = 2 + 99 = … = 101, and there are 100 columns, so 2S = 100 × 101 = 10100.1 mark
3. S = 10100 ÷ 2 = 5050. Yes, the same approach works for any number of terms.½ mark

Check: Formula n(n + 1)/2 with n = 100: 100 × 101 ÷ 2 = 5050 ✓.

Answer to write in the exam

Common mistakes that cost marks

• Using 100 × 100 ÷ 2 = 5000. Each pair adds to 101, not 100.
• Counting 50 pairs but then also halving: either use 50 pairs of 101 (= 5050) or 100 columns of 101 halved (= 5050), not both.
• Writing 2S = 10100 as the answer.

How this can come in the exam

Try one yourself

Use the same approach to find 1 + 2 + 3 + … + 40.

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