The nth Term from the Sum

Two sums, one term apart

Sₙ is the sum of the first n terms of a sequence: Sₙ = u₁ + u₂ + … + uₙ. The sum of the first n − 1 terms, Sₙ₋₁, is the same sum without its last term.

So subtracting one from the other cancels every term but the last: uₙ = Sₙ − Sₙ₋₁. With four terms, S₄ − S₃ = u₄.

From a formula for Sₙ

Suppose Sₙ = n² + 2n. To get Sₙ₋₁, replace every n with n − 1: Sₙ₋₁ = (n − 1)² + 2(n − 1) = n² − 2n + 1 + 2n − 2 = n² − 1.

Then uₙ = (n² + 2n) − (n² − 1) = 2n + 1. Check it by adding terms: the first four are 3, 5, 7 and 9, which add to 24, and S₄ = 16 + 8 = 24.

The terms go up by 2 each time, so this sequence is arithmetic. That always happens when Sₙ = an² + bn: then uₙ = (2n − 1)a + b, which goes up by 2a each step.

The first term

uₙ = Sₙ − Sₙ₋₁ needs a sum of n − 1 terms, so it only makes sense for n ≥ 2. The first term is found on its own: u₁ is the first term, which is the sum of one term, so u₁ = S₁.

Usually the formula still gives the right u₁, but not always. With Sₙ = n² + 2n + 1 the constant 1 cancels in Sₙ − Sₙ₋₁, so uₙ = 2n + 1 for n ≥ 2. But u₁ = S₁ = 1 + 2 + 1 = 4, not 3. The sequence is 4, 5, 7, 9, …, and it is not arithmetic, because its first step is 1.

A geometric example

Take Sₙ = 3ⁿ − 1. Then Sₙ₋₁ = 3ⁿ⁻¹ − 1, and uₙ = 3ⁿ − 3ⁿ⁻¹ = 3 × 3ⁿ⁻¹ − 3ⁿ⁻¹ = 2 × 3ⁿ⁻¹ for n ≥ 2.

Check the first term separately: u₁ = S₁ = 3 − 1 = 2, and 2 × 3⁰ = 2 as well, so the formula holds for every n. The terms 2, 6, 18, … form a geometric sequence with first term 2 and ratio 3.

The usual slips

Differentiating Sₙ: n counts terms, and the change from one sum to the next is a subtraction, Sₙ − Sₙ₋₁, not a derivative.

Giving Sₙ₋₁, or Sₖ itself, as the term: both are sums of many terms.

Using the formula for u₁ without checking it against S₁, which goes wrong whenever Sₙ has a constant term that is not 0.