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Hi James. Start by checking the first first values of n: f(1) = 1 ≤ 2^{11} = 2^{0} = 1. TRUE. So we know it is true for all n up to and including n = 3. We've shown it is certainly true that f(k) ≤ 2^{k1} for k = 3, because we tested that explicitly. If we can prove that given this, then f(k+1) ≤ 2^{k}, then it must work for all positive values of k. In other words, In inequalities, we can add the same quantity to both sides and still have a valid inequality. Therefore, let's add something to both sides. f(k) + f(k1) ≤ 2^{k1} + f(k1) But since f(k1) ≤ 2^{[k1]  1} (we showed that is true for k = 3 at the outset), then we have a third part to the inequality we can add to the end: f(k) + f(k1) ≤ 2^{k1} + f(k1) ≤ 2^{k1} + 2^{[k1]  1} By definition though, the left side equals f(k+1), though, right? And let's just drop the middle section which has outlived its usefulness. Thus we have f(k+1) ≤ 2^{k1} + 2^{k2}. Can you simplify the right side? Stephen La Rocque  


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