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?

Step-by-step solution

Idea: Put the value of n into n(n + 1) ÷ 2. Halve whichever of n and n + 1 is even first, to keep the numbers small.

1. S₂₀ = 20 × 212 = 10 × 21 = 210.1 mark
2. S₅₀ = 50 × 512 = 25 × 51 = 1275.1 mark
3. S₁₀₀₀ = 1000 × 10012 = 500 × 1001 = 500500.1 mark

Check: S₂₀ by pairing: 10 pairs of (1 + 20) = 10 × 21 = 210 ✓.

Answer to write in the exam

Common mistakes that cost marks

• Forgetting to divide by 2 (getting 420, 2550, 1001000).
• Using n(n − 1) ÷ 2 (getting 190 for n = 20).
• Arithmetic slip in 25 × 51: 25 × 50 + 25 = 1275.

How this can come in the exam

Try one yourself

Find S₂₅ and S₂₀₀.

More questions like this

• 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.
• Find the nth term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP.
• An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.
  (Hint: If ‘a’ is the first term and ‘d’ the common difference, then we arrive at the equations a + 2d = 12 and a + 49d = 106. Solve this pair of linear equations for ‘a’ and ‘d’.)

All Sequences and progressions questions · All maths questions