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Question Number 176070 by doline last updated on 11/Sep/22

demontrer par recurrence que pour tout n>0 appartenent a l ensemble des entier naturel 3^(2n−2^(n ) ) est multiple de 7

$${demontrer}\:{par}\:{recurrence}\:{que}\:{pour}\:{tout}\:{n}>\mathrm{0}\:{appartenent}\:{a}\:{l}\:{ensemble}\:{des}\:{entier}\:{naturel}\:\mathrm{3}^{\mathrm{2}{n}−\mathrm{2}^{{n}\:} } {est}\:{multiple}\:{de}\:\mathrm{7} \\ $$

Commented bysom(math1967) last updated on 11/Sep/22

3^(2n) −2^n p(1)=3^2 −2^1 =7 let true for p(m) ∴3^(2m) −2^m =7k 3^(2m) =7k+2^m p(m+1)=3^(2m+2) −2^(m+1) =3^(2m) ×9−2^(m+1) =(7k+2^m )×9−2^(m+1) =63k +2^m ×9−2^(m+1) =63k +2^m (9−2) =7(9k+2^m ) ∴ true for p(m+1) ∴ 3^(2n) −2^n multiple de 7 n∈N

$$\mathrm{3}^{\mathrm{2}{n}} −\mathrm{2}^{{n}} \\ $$ $${p}\left(\mathrm{1}\right)=\mathrm{3}^{\mathrm{2}} −\mathrm{2}^{\mathrm{1}} =\mathrm{7} \\ $$ $${let}\:{true}\:{for}\:{p}\left({m}\right) \\ $$ $$\:\therefore\mathrm{3}^{\mathrm{2}{m}} −\mathrm{2}^{{m}} =\mathrm{7}{k} \\ $$ $$\:\mathrm{3}^{\mathrm{2}{m}} =\mathrm{7}{k}+\mathrm{2}^{{m}} \\ $$ $${p}\left({m}+\mathrm{1}\right)=\mathrm{3}^{\mathrm{2}{m}+\mathrm{2}} −\mathrm{2}^{{m}+\mathrm{1}} \\ $$ $$=\mathrm{3}^{\mathrm{2}{m}} ×\mathrm{9}−\mathrm{2}^{{m}+\mathrm{1}} \\ $$ $$=\left(\mathrm{7}{k}+\mathrm{2}^{{m}} \right)×\mathrm{9}−\mathrm{2}^{{m}+\mathrm{1}} \\ $$ $$=\mathrm{63}{k}\:+\mathrm{2}^{{m}} ×\mathrm{9}−\mathrm{2}^{{m}+\mathrm{1}} \\ $$ $$=\mathrm{63}{k}\:+\mathrm{2}^{{m}} \left(\mathrm{9}−\mathrm{2}\right) \\ $$ $$=\mathrm{7}\left(\mathrm{9}{k}+\mathrm{2}^{{m}} \right) \\ $$ $$\therefore\:{true}\:{for}\:{p}\left({m}+\mathrm{1}\right) \\ $$ $$\therefore\:\mathrm{3}^{\mathrm{2}{n}} −\mathrm{2}^{{n}} \:\:{multiple}\:{de}\:\mathrm{7}\:\:{n}\in{N} \\ $$

Commented bysom(math1967) last updated on 11/Sep/22

3^(2n−2^n ) or 3^(2n) −2^n ?

$$\:\mathrm{3}^{\mathrm{2}{n}−\mathrm{2}^{{n}} } \:{or}\:\mathrm{3}^{\mathrm{2}{n}} −\mathrm{2}^{{n}} \:\:? \\ $$

Commented bydoline last updated on 11/Sep/22

multiple de 7

$${multiple}\:{de}\:\mathrm{7} \\ $$

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