Question

a) Arătați că numărul A = 2ⁿ⁺² · 3ⁿ⁺³ + 2ⁿ⁺³ · 3ⁿ⁺² divizibil cu 180, oricare n număr natural.
b) Arătați că numărul B = 2ⁿ · 3ⁿ⁺¹ + 6ⁿ · 5 + 2ⁿ⁺¹ · 3ⁿ⁺¹ divizibil cu 21, oricare n număr natural nenul.
c) Arătați că numărul C = 2¹ + 2² + 2³ + ... + 2²⁰¹³ divizibil cu 7.

Answer 1

......................................................

Answer 2

a)

$A=2^{n+2}*3^{n+3}+2^{n+3}*3^{n+2}$

$A=2^{n}*2^{2}*3^{n}*3^{3}+2^{n}*2^{3}*3^{n}*3^{2}$

$A=(2*3)^{n}*4*27+(2*3)^{n}*8*9$

$A=6^{n}*102+6^{n}*72$

$A=6^{n}(102+72)$

$A=6^{n}*\bf 108$

$=>~numarul~A~este~divizibil~cu~108$


b)
$B=2^{n}*3^{n+1}+6^{n}*5+2^{n+1}*3^{n+1}$

$B=2^{n}*3^{n}*3+6^{n}*5+2^{n}*2*3^{n}*3$

$B=(2*3)^{n}*3+(2*3)^{n}*5+(2*3)^{n}*6$

$B=(2*3)^{n}(3+5+6)$

$B=2^{n}*3^{n}*14$

$B=2^{n}*3^{n-1}*3*7*2$

$B=2^{n}*3^{n-1}*2*\bf 21$

$=>~B~este~divizibil~cu~21$


c)
$C=2^{1}+2^{2}+2^{3}+2^{4}+2^{5}+2^{6}+2^{7}+2^{8}+2^{9}...+2^{2011}+2^{2012}+2^{2013}$

$C=1*(2^{1}+2^{2}+2^{3})+2^{3}(2^{1}+2^{2}+2^{3})+2^{6}(2^{1}+2^{2}+2^{3})+...+2^{2010}(2^{1}+2^{2}+2^{3})$

$C=1*(2+4+8)+2^{3}(2+4+8)+2^{6}(2+4+8)+...+2^{2010}(2+4+8)$

$C=1*14+2^{3}*14+2^{6}*14+...+2^{2010}*14$

$C=14(1+2^{3}+2^{6}+2^{9}+...+2^{2010})$

$C=7*2*(1+2^{3}+2^{6}+2^{9}+...+2^{2010})$

$=>~C~este~divizibil~cu~7$