Question

[tex]Aratati\ ca\ pentru\ orice\ numar\ natural\ n\ avem:\\
a)(3^n-1)(10*25^n+9*6^n)\ se\ divide\ cu\ 38.\\
b)3*5^{2n+1}+2^{3n+1}\ se\ divide\ cu\ 17.\\
c)2^{2^{n+2}}}-1\ se\ divide\ cu\ 15.\\
Prin\ inductie\ matematica\ va\ rog.[/tex]

Answer 1

$a)~Pentru~n=0~expresia~este~egala~cu~0,~deci~propozitia~este~ \\ \\ adevarata.~Pentru~n=1~este~de~asemenea~adevarata~(verificare~ \\ \\ prin~calcul).~Presupunem~ca~propozitia~este~adevarata~pentru \\ \\ k \in N^*~si~demonstram~ca~este~adevarata~pentru~k+1. \\ \\ 3^n-1~este~par~ \forall~ n \in N^*.~Deci~vom~demonstra~ca~a~doua~paranteza \\ \\ este~divizibila~cu~19. \\ \\ 10 \cdot 25^k+9 \cdot 6^k~ \vdots~19.$

$10 \cdot 25^{k+1}+9 \cdot 6^{k+1}=250 \cdot 25^k +54 \cdot 6^k = 6(10 \cdot 25^k+9 \cdot 6^k)+ \\ \\ +190 \cdot 25^k=M_{38}+38 \cdot 5 \cdot 25^k~ \vdots ~38. \\ \\ b)~n=0,~adevarat.~Pres.~ca.~prop.~este~adev.~pentru~n=k~(k \in N)~si \\ \\ dem.~ca~este~adev.~si~pentru~n=k+1. \\ \\ 3 \cdot 5^{2k+1} +2 ^{3k+1} ~ \vdots~17. \\ \\ 3 \cdot 5^{2k+3}+ 2^{3k+4}=75 \cdot 5^{2k+1}+8 \cdot 2^{3k+1}=8(3 \cdot 5^{2k+1}+2^{3k+1}) \\ \\ + 51 \cdot 5^{2k+1}=M_{17}+17 \cdot 3 \cdot 5^{2k+1} ~ \vdots~17.$

$c)~n=0,~adevarat.~Aceeasi~presupunere. \\ \\ 2^{2^{k+3}}-1=2^{2^{k+2} \cdot 2 }-1=\Big (2^{2^{k+2}} \Big)^2-1= \underbrace{(2^{2^{k+2}}-1)}(2^{2^{k+2}}+1)~ \vdots~15. \\ \\ Am~folosit:~a^2-b^2=(a-b)(a+b).$