Prove-that-i-2-n-gt-n-2-for-all-integral-values-of-n-5-ii-n-gt-3-n-1-for-all-integral-values-of-n-5-

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 by RasheedAhmad last updated on 26/Nov/15

N i c e!