Terms from the sums
Show on a big screenA hall's rows hold 16, 19, 22, 25, 28 seats. The sums so far are 16, 35, 57, 82, 110. These are running totals. Subtract each total from the next: 19, 22, 25, 28. These are the list's terms after the first.
- The running totals are bars: S₁ = 16, S₂ = 35, S₃ = 57, S₄ = 82.
- Bar S₁ has no bar before it. So a₁ = S₁ = 16.
- Cut S₂ level with S₁. The top piece is a₂ = S₂ − S₁ = 35 − 16 = 19.
- Cut S₃ level with S₂: a₃ = 57 − 35 = 22. Cut S₄ level with S₃: a₄ = 82 − 57 = 25.
Rule: a₁ = S₁, since nothing comes before the first term. Then aₙ = Sₙ − Sₙ₋₁ for n = 2, 3, 4, and so on.
Here Sₙ is the total of the first n terms. Sₙ₋₁ is the total of the first n − 1 terms. Sₙ has one more term than Sₙ₋₁. So Sₙ − Sₙ₋₁ is the nth term. As bars, Sₙ₋₁ is the part of bar Sₙ below the cut. The top piece is the nth term.
Sometimes Sₙ is given as a formula. Take Sₙ = 2n² + 3n:
| n | Sₙ = 2n² + 3n | aₙ = Sₙ − Sₙ₋₁ |
|---|---|---|
| 1 | 5 | 5 = S₁ |
| 2 | 14 | 14 − 5 = 9 |
| 3 | 27 | 27 − 14 = 13 |
| 4 | 44 | 44 − 27 = 17 |
The list reads 5, 9, 13, 17. The steps are all 4 so far. Now do the same subtraction with letters. Sₙ₋₁ = 2(n − 1)² + 3(n − 1) = 2n² − 4n + 2 + 3n − 3 = 2n² − n − 1. So aₙ = (2n² + 3n) − (2n² − n − 1) = 4n + 1. The step is 4 for every n. So this Sₙ describes an AP.
Drag each bar apart. Cut it level with the bar before. The top piece is a term.
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Try the lab to finish this lesson.