Numerical answer
A function g satisfies g(x) = 2·g(x−1) − 1 for all integers x, with g(1) = 3. Find g(6).
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Answer: 65
Define h(x) = g(x) − 1. Then h(x) = g(x) − 1 = 2g(x−1) − 1 − 1 = 2(g(x−1) − 1) = 2h(x−1), so h(x) = 2h(x−1), a geometric recursion. h(1) = g(1) − 1 = 2, so h(n) = 2ⁿ for n ≥ 1. Thus g(n) = 2ⁿ + 1. g(6) = 2⁶ + 1 = 64 + 1 = 65.
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