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A function f is defined on the positive integers by f(1) = 1, and for each integer n greater than 1, f(n) = f(n−1) + n if n is even, and f(n) = f(n−1) − n if n is odd. What is f(50)?
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Correct answer
B
Running all 50 steps is the slow road; the design rewards computing just enough terms to trust a pattern at the even arguments.
Walk the start: f(1) = 1; f(2) = 1 + 2 = 3; f(3) = 3 − 3 = 0; f(4) = 0 + 4 = 4; f(5) = 4 − 5 = −1; f(6) = −1 + 6 = 5; f(7) = 5 − 7 = −2; f(8) = −2 + 8 = 6.
Read off the even arguments: f(2) = 3, f(4) = 4, f(6) = 5, f(8) = 6. Each even step nets +1, because going two arguments forward you subtract an odd term and then add the next (even) term: −(2k−1) + 2k = +1. So f(2k) = k + 2. For n = 50 you have k = 25 and f(50) = 25 + 2 = 27.
The principle: find the rate, then anchor it at a known term before you jump. The tempting move, once you see the +1-per-even-step rate, is to write f(50) = 25 with no offset, choice A, the rate is right but the anchor is wrong, and checking the formula against f(2) = 3 exposes the needed + 2. Choice C uses the whole value 3 as the offset (25 + 3 = 28). Choices D and E attach the offset to the argument 50 itself (50 + 2 = 52, 50 + 3 = 53) rather than to k = 25. So f(50) = 27, choice B.
Is there another way to approach this question? Yes, sum the signed terms directly. By definition f(50) = 1 + (2 − 3 + 4 − 5 + 6 − 7 + ⋯ + 48 − 49 + 50). Group consecutive even-then-odd pairs: (2 − 3), (4 − 5), …, (48 − 49) each equal −1, and there are 24 such pairs, with the term 50 left unpaired. So the sum from 2 to 50 is 24·(−1) + 50 = 26, and f(50) = 1 + 26 = 27. This direct count confirms the pattern result.
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