Prove that: (i) 2^n >n^2 for all integral values of n≥5 (ii) n!>3^(n−1) ,for all integral values of n≥5


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Question Number 2750 by RasheedAhmad last updated on 26/Nov/15

$P r o v e t h a t :$ $(i) 2^{n} > n^{2} f o r a l l i n t e g r a l v a l u e s$ $o f n ⩾ 5$ $(i i) n! > 3^{n - 1}, f o r a l l i n t e g r a l v a l u e s$ $o f n ⩾ 5$
	Answered by Yozzi last updated on 26/Nov/15

	$(S k e l e t o n s o f p r o o f s)$ $(i) L e t P (n) : 2^{n} > n^{2}, \forall n ⩾ 5, n \in Z .$ $P (5) : 2^{5} = 32 > 25 = 5^{2}$ $\Rightarrow P (n) t r u e f o r n = 5 .$ $S u p p p o s e P (n) t r u e f o r n = k (k ⩾ 5) :$ $2^{k} > k^{2}$ $P (k + 1) : 2^{k} > k^{2}$ $\times 2 : 2^{k + 1} > 2 k^{2}$ $2^{k + 1} > k^{2} + k^{2} - 2 k + 1 + 2 k - 1$ $2^{k + 1} > k^{2} - 2 k - 1 + {(k + 1)}^{2}$ $2^{k + 1} > {(k - 1)}^{2} - 2 + {(k + 1)}^{2}$ $∵ k ⩾ 5 \Rightarrow k - 1 ⩾ 4 \Rightarrow {(k - 1)}^{2} ⩾ 16$ ${(k - 1)}^{2} - 2 ⩾ 14 > 0$ $∴ {(k - 1)}^{2} - 2 + {(k + 1)}^{2} > {(k + 1)}^{2}$ $H e n c e, 2^{k + 1} > {(k + 1)}^{2} .$ $∴ P (k) \Rightarrow P (k + 1)$ $∵ P (5) i s t r u e, b y P . M . I, 2^{n} > n^{2}$ $\forall n ⩾ 5, n \in Z .$ $(i i) L e t P (n) : n! > 3^{n - 1}, n ⩾ 5, n \in Z .$ $P (5) : 5! = 120 > 3^{4} = 81$ $∴ P (n) t r u e f o r n = 5 .$ $S u p p o s e P (n) t r u e f o r n = k (k ⩾ 5) :$ $k! > 3^{k - 1}$ $P (k + 1) : k! > 3^{k - 1}$ $\times (k + 1 > 0) : (k + 1)! > (k + 1) 3^{k - 1}$ $∵ k ⩾ 5 \Rightarrow k + 1 ⩾ 6 \Rightarrow (k + 1) 3^{k - 1} ⩾ 6 \times 3^{k - 1}$ $(k + 1) 3^{k - 1} ⩾ 2 \times 3^{k} > 3^{k}$ $H e n c e, (k + 1)! > 3^{[k + 1] - 1} .$ $∴ P (k) \Rightarrow P (k + 1)$ $S i n c e P (5) i s t r u e, b y P . M . I,$ $k! > 3^{k - 1} \forall n ⩾ 5, n \in Z .$
	Commented byRasheedAhmad last updated on 26/Nov/15

	$N i c e!$
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