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Cum se rezolva exercitiul 2×(1+3+3^2+3^3+.....+3^49)=?

Răspuns :

2(3^0+3^1+3^2+....+3^49)=
   |__________________|-progresie geometrica cu 50 termeni,ratia=3
suma unei progresii geometrice:Sn=a1(r^n-1)/r-1
S49=1(3^50-1)/2
2* (3^50-1)/2=3^50-1 rezultata final
2×(1+3+3²+...+3⁴⁹)=2 x (3⁵⁰-1)/(3-1) = 3⁵⁰ - 1