Alexpaveliuc12parykfask
a fost răspuns

Aratati ca nr. C = 2¹+2²+2³+.......2²⁰¹³ divizibil cu 7
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Răspuns :

Kawaiimathask

C=2^1+2^2+2^3+...+2^2013

C=2^1(1+2^1+2^2)+2^4(1+2^1+2^2)+...+2^2011(1+2^1+2^2)

C=2*(1+2+4)+2^4(1+2+4)+...+2^2011(1+2+4)

C=2*7+2^4*7+...+2^2011*7

C=7(2+2^4+...+2^2011) ⋮ 7