Sum of n terms

A famous story. When Gauss (from Chapter 1) was just 10 years old, he was asked to add 1 + 2 + 3 + … + 100. He answered at once: 5050. How?

He wrote the sum forwards and backwards, one under the other:

S = 1 + 2 + 3 + … + 99 + 100 S = 100 + 99 + 98 + … + 2 + 1

Adding each column gives 101 every time, and there are 100 columns. So 2S = 100 × 101, and S = 5050.

The same trick works for any AP. Each column adds to (first term + last term), and there are n columns. So:

(The second comes from the first, because l = a + (n – 1)d.)

Sₙ means the first n terms added up. S₁ is just the first term, and S₂ = a₁ + a₂. Careful: S₂ is not a₂. To get back one term, take one sum from the next: aₙ = Sₙ – Sₙ₋₁.

Adding 1 to n: 1 + 2 + … + n = n(n + 1)/2. Check with n = 100: 100 × 101/2 = 5050 ✓.

Two answers can both be right. For 24, 21, 18, …, the sum is 78 after 4 terms and again after 13 terms. That's because terms 5 to 13 (12, 9, 6, 3, 0, –3, –6, –9, –12) add up to 0. Keep both answers.