# Prove the following by the principle of mathematical induction. 2n?>?n2, where?n?is a positive integer such that?n?> 4. Solution: Let the given statement be P(n), i.e., P(n) : 2n?>?n2?where?n?> 4 For?n?= 5, 25?= 32 and 52?= 25 ?25?> 52 Thus, P(n) is true for?n?= 5. Let P(n) be true for?n?=?k, i.e., 2k?>?k2?? (1) Now, we have to prove that P(k? 1) is true whenever P(k) is true, i.e. we have to prove that 2k? 1?> (k? 1)2. From equation (1), we obtain 2k?>?k2 On multiplying both sides with 2, we obtain 2 ? 2k?> 2 ??k2 2k? 1?> 2k2 ?To prove 2k? 1?> (k? 1)2, we only need to prove that 2k2?> (k? 1)2. Let us assume 2k2?> (k? 1)2. ? 2k2?>?k2? 2k? 1 ??k2?> 2k? 1 ??k2?? 2k?? 1 > 0 ? (k?? 1)2?? 2 > 0 ? (k?? 1)2?> 2, which is true as?k?> 4 Hence, our assumption 2k2?> (k? 1)2?is correct and we have 2k? 1?> (k? 1)2. Thus, P(n) is true for?n?=?k? 1. Thus, by the principle of mathematical induction, the given mathematical statement is true for every positive integer?n. ? IN THIS EXAMPLE IT HAS BEEN WRITTEN THAT , To prove 2k? 1?> (k? 1)2, we only need to prove that 2k2?> (k? 1)2. how?2k? 1?> (k? 1)2?=?2k2?> (k? 1)2 I cant understand how so plz explain that

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