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(

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*) + 1f(n) = |
n(n + 1) |
+ 1 = | n^{2} + n + 2 |
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2 | 2 |