with the use of mathematical induction show that n!>2n^3 , ∀n≥6.


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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
	$when n=6 6!=720 2×6^3 =432 n!>2n^3 when n=6 let when n=p, the given statment is true p!>2p^3 we have to prove that (p+1)!>2(p+1)^3 now (p+1)p!−2(p+1)^3 (p+1){p!−2(p+1)^2 } let assume p!=2p^3 +δ→δ is positive value (p+1){2p^3 +δ−2(p+1)^2 } (p+1)[2{(p^3 −(p+1)^2 }+δ] →is positive because when p≥6 p^3 >(p+1)^2 pls check$
	$w h e n n = 6$ $6! = 720 2 \times 6^{3} = 432 n! > 2 n^{3} w h e n n = 6$ $l e t w h e n n = p, t h e g i v e n s t a t m e n t i s t r u e$ $p! > 2 p^{3}$ $w e h a v e t o p r o v e t h a t$ $(p + 1)! > 2 {(p + 1)}^{3}$ $n o w$ $(p + 1) p! - 2 {(p + 1)}^{3}$ $(p + 1) {p! - 2 {(p + 1)}^{2}}$ $l e t a s s u m e p! = 2 p^{3} + δ \to δ i s p o s i t i v e v a l u e$ $(p + 1) {2 p^{3} + δ - 2 {(p + 1)}^{2}}$ $(p + 1) [2 {(p^{3} - {(p + 1)}^{2}} + δ] \to i s p o s i t i v e$ $b e c a u s e w h e n p ⩾ 6$ $p^{3} > {(p + 1)}^{2} p l s c h e c k$
	Answered by MWSuSon last updated on 25/Sep/20
	$Base case(n=6): 6!=720>2×6^3 =432 Inductive step{Goal: (n+1)!>2(n+1)^3 } suppose k≥6 and k!>2k^3 Observe that (k+1)!=(k+1)k! >(k+1)2k^3 =k(2k^3 )+2k^3 =k(2k^3 )+(k^3 +k^3 ) (observe that ∀ k≥6, k^3 ≥6k^2 >6k+2) >k(2k^3 )+(6k^2 +6k+2) >2k^3 +6k^2 +6k+2 =2(k+1)^3 Therefore n!>2n^3 (verify if this is logical)$
	$Base case (n = 6) : 6! = 720 > 2 \times 6^{3} = 432$ $Inductive step {Goal : (n + 1)! > 2 {(n + 1)}^{3}}$ $suppose k ⩾ 6 and k! > {2 k}^{3}$ $Observe that (k + 1)! = (k + 1) k!$ $> (k + 1) {2 k}^{3}$ $= k ({2 k}^{3}) + {2 k}^{3}$ $= k ({2 k}^{3}) + (k^{3} + k^{3})$ $(observe that \forall k ⩾ 6, k^{3} ⩾ {6 k}^{2} > 6 k + 2)$ $> k ({2 k}^{3}) + ({6 k}^{2} + 6 k + 2)$ $> {2 k}^{3} + {6 k}^{2} + 6 k + 2$ $= 2 {(k + 1)}^{3}$ $Therefore n! > {2 n}^{3}$ $(verify if this is logical)$
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