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aratati ca 2la1+2la2+2la3+.....+2la100 divizibil cu 3

Răspuns :

(2+2^2)+(2^3+2^4)+...+(2^99+2^100)=

2(1+2)+2^3(1+2)+...+2^99(1+2)=

2*3+(2^3)*3+...+(2^99)*3=

3*(2+2^3+2^5+...+2^99)=m3, m=multiplu de 3, deci se divide cu 3.