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

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

$$\mathrm{with}\:\mathrm{the}\:\mathrm{use}\:\mathrm{of}\:\mathrm{mathematical}\:\mathrm{induction} \\ $$ $$\mathrm{show}\:\mathrm{that}\:\mathrm{n}!>\mathrm{2n}^{\mathrm{3}} ,\:\forall\mathrm{n}\geqslant\mathrm{6}. \\ $$

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

$${when}\:{n}=\mathrm{6} \\ $$ $$\mathrm{6}!=\mathrm{720}\:\:\:\:\mathrm{2}×\mathrm{6}^{\mathrm{3}} =\mathrm{432}\:\:\:{n}!>\mathrm{2}{n}^{\mathrm{3}} \:\:{when}\:{n}=\mathrm{6} \\ $$ $${let}\:{when}\:{n}={p},\:{the}\:{given}\:{statment}\:{is}\:{true} \\ $$ $${p}!>\mathrm{2}{p}^{\mathrm{3}} \\ $$ $${we}\:{have}\:{to}\:{prove}\:{that} \\ $$ $$\left({p}+\mathrm{1}\right)!>\mathrm{2}\left({p}+\mathrm{1}\right)^{\mathrm{3}} \\ $$ $${now} \\ $$ $$\left({p}+\mathrm{1}\right){p}!−\mathrm{2}\left({p}+\mathrm{1}\right)^{\mathrm{3}} \\ $$ $$\left({p}+\mathrm{1}\right)\left\{{p}!−\mathrm{2}\left({p}+\mathrm{1}\right)^{\mathrm{2}} \right\} \\ $$ $${let}\:{assume}\:{p}!=\mathrm{2}{p}^{\mathrm{3}} +\delta\rightarrow\delta\:{is}\:{positive}\:{value} \\ $$ $$\left({p}+\mathrm{1}\right)\left\{\mathrm{2}{p}^{\mathrm{3}} +\delta−\mathrm{2}\left({p}+\mathrm{1}\right)^{\mathrm{2}} \right\} \\ $$ $$\left({p}+\mathrm{1}\right)\left[\mathrm{2}\left\{\left({p}^{\mathrm{3}} −\left({p}+\mathrm{1}\right)^{\mathrm{2}} \right\}+\delta\right]\:\rightarrow{is}\:{positive}\right. \\ $$ $${because}\:{when}\:{p}\geqslant\mathrm{6} \\ $$ $${p}^{\mathrm{3}} >\left({p}+\mathrm{1}\right)^{\mathrm{2}} \:\:\:\:{pls}\:{check} \\ $$ $$ \\ $$

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)

$$\mathrm{Base}\:\mathrm{case}\left(\mathrm{n}=\mathrm{6}\right):\:\mathrm{6}!=\mathrm{720}>\mathrm{2}×\mathrm{6}^{\mathrm{3}} =\mathrm{432} \\ $$ $$\mathrm{Inductive}\:\mathrm{step}\left\{\mathrm{Goal}:\:\left(\mathrm{n}+\mathrm{1}\right)!>\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} \right\} \\ $$ $$\mathrm{suppose}\:\mathrm{k}\geqslant\mathrm{6}\:\mathrm{and}\:\mathrm{k}!>\mathrm{2k}^{\mathrm{3}} \\ $$ $$\mathrm{Observe}\:\mathrm{that}\:\left(\mathrm{k}+\mathrm{1}\right)!=\left(\mathrm{k}+\mathrm{1}\right)\mathrm{k}! \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:>\left(\mathrm{k}+\mathrm{1}\right)\mathrm{2k}^{\mathrm{3}} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{k}\left(\mathrm{2k}^{\mathrm{3}} \right)+\mathrm{2k}^{\mathrm{3}} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{k}\left(\mathrm{2k}^{\mathrm{3}} \right)+\left(\mathrm{k}^{\mathrm{3}} +\mathrm{k}^{\mathrm{3}} \right)\: \\ $$ $$\left(\mathrm{observe}\:\mathrm{that}\:\forall\:\mathrm{k}\geqslant\mathrm{6},\:\mathrm{k}^{\mathrm{3}} \geqslant\mathrm{6k}^{\mathrm{2}} >\mathrm{6k}+\mathrm{2}\right) \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:>\mathrm{k}\left(\mathrm{2k}^{\mathrm{3}} \right)+\left(\mathrm{6k}^{\mathrm{2}} +\mathrm{6k}+\mathrm{2}\right) \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:>\mathrm{2k}^{\mathrm{3}} +\mathrm{6k}^{\mathrm{2}} +\mathrm{6k}+\mathrm{2} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{3}} \\ $$ $$\mathrm{Therefore}\:\mathrm{n}!>\mathrm{2n}^{\mathrm{3}} \\ $$ $$ \\ $$ $$\left(\mathrm{verify}\:\mathrm{if}\:\mathrm{this}\:\mathrm{is}\:\mathrm{logical}\right) \\ $$

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