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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