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1. We show that 5n? E N(n). Because, for n 2 0, 5n2 > 1 x na, we can take c = 1 and N
1. We show that 5n? E N(n). Because, for n 2 0, 5n2 > 1 x na, we can take c = 1 and N = 0 to obtain our result
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