Question 15

If $$f(n + 1) = \frac{2f(n) + 1}{2}$$ for all natural numbers $$n$$ and $$f(1) = \frac{1}{2}$$, then $$-f(1) + f(2) - f(3) + f(4) - f(5) + \cdots - f(99) + f(100) =$$

Solution

The recurrence is $$f(n + 1) = f(n) + \frac{1}{2}$$ with $$f(1) = \frac{1}{2}$$, so $$f(n) = \frac{n}{2}$$. Grouping the sum in pairs, each pair contributes $$f(2k) - f(2k - 1) = \frac{1}{2}$$. There are 50 such pairs, so the total is $$50 \times \frac{1}{2} = 25$$.

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