with-the-use-of-mathematical-induction-show-that-n-gt-2n-3-n-6-

Question Number 115417 by toa last updated on 25/Sep/20

with the use of mathematical induction

show that n! > {2 n}^{3}, \forall n ⩾ 6 .

Answered by TANMAY PANACEA last updated on 25/Sep/20

w h e n n = 6

6! = 720 2 \times 6^{3} = 432 n! > 2 n^{3} w h e n n = 6

l e t w h e n n = p, t h e g i v e n s t a t m e n t i s t r u e

p! > 2 p^{3}

w e h a v e t o p r o v e t h a t

(p + 1)! > 2 {(p + 1)}^{3}

n o w

(p + 1) p! - 2 {(p + 1)}^{3}

(p + 1) {p! - 2 {(p + 1)}^{2}}

l e t a s s u m e p! = 2 p^{3} + δ \to δ i s p o s i t i v e v a l u e

(p + 1) {2 p^{3} + δ - 2 {(p + 1)}^{2}}

(p + 1) [2 {(p^{3} - {(p + 1)}^{2}} + δ] \to i s p o s i t i v e

b e c a u s e w h e n p ⩾ 6

p^{3} > {(p + 1)}^{2} p l s c h e c k

Answered by MWSuSon last updated on 25/Sep/20

Base case (n = 6) : 6! = 720 > 2 \times 6^{3} = 432

Inductive step {Goal : (n + 1)! > 2 {(n + 1)}^{3}}

suppose k ⩾ 6 and k! > {2 k}^{3}

Observe that (k + 1)! = (k + 1) k!

> (k + 1) {2 k}^{3}

= k ({2 k}^{3}) + {2 k}^{3}

= k ({2 k}^{3}) + (k^{3} + k^{3})

(observe that \forall k ⩾ 6, k^{3} ⩾ {6 k}^{2} > 6 k + 2)

> k ({2 k}^{3}) + ({6 k}^{2} + 6 k + 2)

> {2 k}^{3} + {6 k}^{2} + 6 k + 2

= 2 {(k + 1)}^{3}

Therefore n! > {2 n}^{3}

(verify if this is logical)

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