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Proof-for-the-following-n-2-3-n-2-




Question Number 111954 by Hassen_Timol last updated on 05/Sep/20
Proof for the following :              n! ≥ 2×3^(n−2)
$${P}\mathrm{roof}\:\mathrm{for}\:\mathrm{the}\:\mathrm{following}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{n}!\:\geqslant\:\mathrm{2}×\mathrm{3}^{{n}−\mathrm{2}} \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 05/Sep/20
by recurence  n=2   2!≥2×1 (true) let suppose n!≥2.3^(n−2)   (n+1)! =(n+1)n! ≥(n+1)2.3^(n−2)   we have  2(n+1)3^(n−2) −2 .3^(n+1−2)  =2(n+1)3^(n−2) −2.3^(n−2+1)   =2.3^(n−2) {n+1−3} =2.3^(n−2) (n−2) ≥0 due to n≥2 ⇒  (n+1)!≥2.3^(n+1−2)    the relation is true  at term (n+1)
$$\mathrm{by}\:\mathrm{recurence}\:\:\mathrm{n}=\mathrm{2}\:\:\:\mathrm{2}!\geqslant\mathrm{2}×\mathrm{1}\:\left(\mathrm{true}\right)\:\mathrm{let}\:\mathrm{suppose}\:\mathrm{n}!\geqslant\mathrm{2}.\mathrm{3}^{\mathrm{n}−\mathrm{2}} \\ $$$$\left(\mathrm{n}+\mathrm{1}\right)!\:=\left(\mathrm{n}+\mathrm{1}\right)\mathrm{n}!\:\geqslant\left(\mathrm{n}+\mathrm{1}\right)\mathrm{2}.\mathrm{3}^{\mathrm{n}−\mathrm{2}} \:\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)\mathrm{3}^{\mathrm{n}−\mathrm{2}} −\mathrm{2}\:.\mathrm{3}^{\mathrm{n}+\mathrm{1}−\mathrm{2}} \:=\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)\mathrm{3}^{\mathrm{n}−\mathrm{2}} −\mathrm{2}.\mathrm{3}^{\mathrm{n}−\mathrm{2}+\mathrm{1}} \\ $$$$=\mathrm{2}.\mathrm{3}^{\mathrm{n}−\mathrm{2}} \left\{\mathrm{n}+\mathrm{1}−\mathrm{3}\right\}\:=\mathrm{2}.\mathrm{3}^{\mathrm{n}−\mathrm{2}} \left(\mathrm{n}−\mathrm{2}\right)\:\geqslant\mathrm{0}\:\mathrm{due}\:\mathrm{to}\:\mathrm{n}\geqslant\mathrm{2}\:\Rightarrow \\ $$$$\left(\mathrm{n}+\mathrm{1}\right)!\geqslant\mathrm{2}.\mathrm{3}^{\mathrm{n}+\mathrm{1}−\mathrm{2}} \:\:\:\mathrm{the}\:\mathrm{relation}\:\mathrm{is}\:\mathrm{true}\:\:\mathrm{at}\:\mathrm{term}\:\left(\mathrm{n}+\mathrm{1}\right) \\ $$
Commented by aleks041103 last updated on 05/Sep/20
He used proof by induction
$${He}\:{used}\:{proof}\:{by}\:{induction} \\ $$
Commented by Hassen_Timol last updated on 05/Sep/20
Sorry but I don't understand I think... ��
Commented by mathmax by abdo last updated on 05/Sep/20
 I have used recurrence method...
$$\:\mathrm{I}\:\mathrm{have}\:\mathrm{used}\:\mathrm{recurrence}\:\mathrm{method}… \\ $$

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