[tex] S_{1} =1+2+ 2^{2} + 2^{3} +...+ 2^{2014} \\ \text{Observam ca suma primilor termeni este: } \\
1+2+2^2 = 1+2+4= 7 \\ =\ \textgreater \ ~~\text{Putem grupa toti termenii in grupe de cate 3 termeni.} \\
S_{1} =(1+2+ 2^{2}) + (2^{3} + 2^4+2^5) +(2^6+2^7+2^8)+... \\
S_{1} =1(1+2+ 2^{2}) + 2^3(1 + 2+2^2) +2^6(1+2+2^2)+... \\
S_{1} =1 \times 7 + 2^3 \times 7 +2^6 \times 7+... \\
S_{1} = 7(1+2^3+2^6 +....) ~~ =\ \textgreater \ ~~ Aparent ~S_1~este~ divizibil~cu~7. \\
[/tex]
[tex]\text{Verificam daca avem voie sa grupam termenii llui } S_1, \\ ~in grupe ~de ~3~termeni. \\
\text{Primul termen este }2^0~iar~ultimul~este~2^{2014} \\ =\ \textgreater \ ~~ Sunt~~ 2015 ~termeni. \\
2015~nu~este~divizibil~cu~3. \\
=\ \textgreater \ ~ \text{Nu putem face gruparea de mai sus} \\
=\ \textgreater \ \boxed{S_1 ~~ nu~este~divizibil~cu~7 }[/tex]
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[tex] S_{2} =1+3+ 3^{2} + 3^{3} +...+ 3^{2015} \\
\text{Obsevam ca suma primilor 4 termeni = 40} \\
1+3+ 3^{2} + 3^{3} = 1+3+9+27 = 40 \\
=\ \textgreater \ ~~\text{Vom gruma sirul de termeni in grupe de cate 4 termeni} \\
\text{Verificam daca avem voie: } \\ Primul ~termen = 1 = 3^0 ~~su ~ultimul = 3^{2015} \\
=\ \textgreater \ ~~In~total~avem~2016~termeni. \\
2016~este~divizibil~cu~4~adica~~~\boxed{2016 \;\vdots\;4} \\ =\ \textgreater \ ~~\text{Avem voie sa grupam termenii in grupe de cate 4 termeni.} \\
[/tex]
[tex]S_2=1+3+ 3^{2} + 3^{3} +3^4+3^5+3^6+3^7+3^8+3^9+3^{10}+3^{11}+... \\ ...+3^{2012}+3^{2013}+3^{2014}+3^{2015}= \\ \\
=(1+3+ 3^{2} + 3^{3}) +(3^4+3^5+3^6+3^7)+(3^8+3^9+3^{10}+3^{11})+... \\ ...+(3^{2012}+3^{2013}+3^{2014}+3^{2015})= \\ =1(1+3+ 3^{2} + 3^{3}) +3^4(1+3+3^2+3^3)+3^8(3+3+3^2+3^3)+... \\ ...+3^{2012}(1+3 +3^2 +3^3)= \\
=(1+3+ 3^{2} + 3^{3})(1+3^4+3^8+...+3^{2012})= \\
= (1+3+ 9 + 27)(1+3^4+3^8+...+3^{2012})= \\
= \boxed{40(1+3^4+3^8+...+3^{2012}) ~\vdots~40} ~(Este~divizibil~cu~40)\\ cctd[/tex]
