If f(1) = 2 and f(n) = f(n – 1) + n, then what’s a formula for f(n)?
f(1) = 2
f(2) = f(1) + 2 = 2 + 2 = 4
f(3) = f(2) + 3 = 4 + 3 = 7
f(4) = f(3) + 4 = 7 + 4 = 11
f(5) = f(4) + 5 = 11 + 5 = 16
f(2) = f(1) + 2 = 2 + 2 = 4
f(3) = f(2) + 3 = 4 + 3 = 7
f(4) = f(3) + 4 = 7 + 4 = 11
f(5) = f(4) + 5 = 11 + 5 = 16
Hmm. This has the look of 1 + 2 + 3 + 4 + … + n. Hah!
f(1) = 2 = 1 + 1
f(2) = 4 = (1 + 2) + 1
f(3) = 7 = (1 + 2 + 3) + 1
f(4) = 11 = (1 + 2 + 3 + 4) + 1
f(5) = 16 = (1 + 2 + 3 + 4 + 5) + 1
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f(n) = (1 + 2 + 3 + … + n) + 1
f(2) = 4 = (1 + 2) + 1
f(3) = 7 = (1 + 2 + 3) + 1
f(4) = 11 = (1 + 2 + 3 + 4) + 1
f(5) = 16 = (1 + 2 + 3 + 4 + 5) + 1
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f(n) = (1 + 2 + 3 + … + n) + 1
| f(n) = | n(n + 1) | + 1 = | n2 + n + 2 | . |
| 2 | 2 |