Question Number 115417 by toa last updated on 25/Sep/20
$$\mathrm{with}\:\mathrm{the}\:\mathrm{use}\:\mathrm{of}\:\mathrm{mathematical}\:\mathrm{induction} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{n}!>\mathrm{2n}^{\mathrm{3}} ,\:\forall\mathrm{n}\geqslant\mathrm{6}. \\ $$
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](https://www.tinkutara.com/question/Q115432.png)
$${when}\:{n}=\mathrm{6} \\ $$$$\mathrm{6}!=\mathrm{720}\:\:\:\:\mathrm{2}×\mathrm{6}^{\mathrm{3}} =\mathrm{432}\:\:\:{n}!>\mathrm{2}{n}^{\mathrm{3}} \:\:{when}\:{n}=\mathrm{6} \\ $$$${let}\:{when}\:{n}={p},\:{the}\:{given}\:{statment}\:{is}\:{true} \\ $$$${p}!>\mathrm{2}{p}^{\mathrm{3}} \\ $$$${we}\:{have}\:{to}\:{prove}\:{that} \\ $$$$\left({p}+\mathrm{1}\right)!>\mathrm{2}\left({p}+\mathrm{1}\right)^{\mathrm{3}} \\ $$$${now} \\ $$$$\left({p}+\mathrm{1}\right){p}!−\mathrm{2}\left({p}+\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\left({p}+\mathrm{1}\right)\left\{{p}!−\mathrm{2}\left({p}+\mathrm{1}\right)^{\mathrm{2}} \right\} \\ $$$${let}\:{assume}\:{p}!=\mathrm{2}{p}^{\mathrm{3}} +\delta\rightarrow\delta\:{is}\:{positive}\:{value} \\ $$$$\left({p}+\mathrm{1}\right)\left\{\mathrm{2}{p}^{\mathrm{3}} +\delta−\mathrm{2}\left({p}+\mathrm{1}\right)^{\mathrm{2}} \right\} \\ $$$$\left({p}+\mathrm{1}\right)\left[\mathrm{2}\left\{\left({p}^{\mathrm{3}} −\left({p}+\mathrm{1}\right)^{\mathrm{2}} \right\}+\delta\right]\:\rightarrow{is}\:{positive}\right. \\ $$$${because}\:{when}\:{p}\geqslant\mathrm{6} \\ $$$${p}^{\mathrm{3}} >\left({p}+\mathrm{1}\right)^{\mathrm{2}} \:\:\:\:{pls}\:{check} \\ $$$$ \\ $$
Answered by MWSuSon last updated on 25/Sep/20
$$\mathrm{Base}\:\mathrm{case}\left(\mathrm{n}=\mathrm{6}\right):\:\mathrm{6}!=\mathrm{720}>\mathrm{2}×\mathrm{6}^{\mathrm{3}} =\mathrm{432} \\ $$$$\mathrm{Inductive}\:\mathrm{step}\left\{\mathrm{Goal}:\:\left(\mathrm{n}+\mathrm{1}\right)!>\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} \right\} \\ $$$$\mathrm{suppose}\:\mathrm{k}\geqslant\mathrm{6}\:\mathrm{and}\:\mathrm{k}!>\mathrm{2k}^{\mathrm{3}} \\ $$$$\mathrm{Observe}\:\mathrm{that}\:\left(\mathrm{k}+\mathrm{1}\right)!=\left(\mathrm{k}+\mathrm{1}\right)\mathrm{k}! \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:>\left(\mathrm{k}+\mathrm{1}\right)\mathrm{2k}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{k}\left(\mathrm{2k}^{\mathrm{3}} \right)+\mathrm{2k}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{k}\left(\mathrm{2k}^{\mathrm{3}} \right)+\left(\mathrm{k}^{\mathrm{3}} +\mathrm{k}^{\mathrm{3}} \right)\: \\ $$$$\left(\mathrm{observe}\:\mathrm{that}\:\forall\:\mathrm{k}\geqslant\mathrm{6},\:\mathrm{k}^{\mathrm{3}} \geqslant\mathrm{6k}^{\mathrm{2}} >\mathrm{6k}+\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:>\mathrm{k}\left(\mathrm{2k}^{\mathrm{3}} \right)+\left(\mathrm{6k}^{\mathrm{2}} +\mathrm{6k}+\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:>\mathrm{2k}^{\mathrm{3}} +\mathrm{6k}^{\mathrm{2}} +\mathrm{6k}+\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\mathrm{Therefore}\:\mathrm{n}!>\mathrm{2n}^{\mathrm{3}} \\ $$$$ \\ $$$$\left(\mathrm{verify}\:\mathrm{if}\:\mathrm{this}\:\mathrm{is}\:\mathrm{logical}\right) \\ $$