Question

demonstrati ca urmatoarele numere sunt divizibile cu 13
a)....il stiu
b)B=3+3^2+3^3+...+3^2015


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Answer 1

Răspuns: Ai demonstrația mai jos

Explicație pas cu pas:

Salutare !

$\large \bf B= 3+3^2+3^3+...+3^{2015}$

$\large \bf B= (3^{1}+3^{2}+3^{3})+(3^{4}+3^{5}+3^{6})+...+(3^{2013}+3^{2014}+3^{2015})$

$\large \bf B=3^{1}\cdot (3^{0}+3^{1}+3^{2})+3^{4}\cdot(3^{0}+3^{1}+3^{2})+...+3^{2013}\cdot(3^{0}+3^{1}+3^{2})$

$\large \bf B=3^{1}\cdot (1+3+9)+3^{4}\cdot(1+3+9)+...+3^{2013}\cdot(1+3+9)$

$\large \bf B=3^{1}\cdot 13+3^{4}\cdot 13+...+3^{2013}\cdot 13$

$\color{red}\large \boxed{\bf B=13 \cdot (3^{1}+3^{4}+...+3^{2013})\implies B~\vdots~13}$

#copaceibrainly    

Answer 2

$\it 3+3^2+3^3=3(1+3+9)=3\cdot13\\ \\ 3^4+3^5+3^6=3^4(1+3+9)=3^4\cdot13\\ \\ 3^{2013}+3^{2014}+3^{2015}=3^{2013}(1+3+9)=3^{2013}\cdot13$

$\it B=13(3+3^4+3^7+\ ...\ +3^{2013}) \Rightarrow B\ \vdots\ 13$