Question

imi raspunde repede, va rog!​

Answer 1

Ai rezolvarea ataşată. Mult succes!

Answer 2

$B=1+3^1+3^2+...+3^{116} \ \vdots \ 13 \\ 13=1+3+3^2=4+9 \\ \Rightarrow \ B=1+3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+...+3^{115}+3^{116}= \\ =1+3+3^2+3^3\cdot(1+3+3^2)+3^6\cdot (1+3+3^2)+...3^{114}+3^{115}+3^{116} \\ In \ total \ sunt \ 117 \ termeni, \ grupati \ cate \ 3: \\ \frac{117}{3}=39 \ \Longrightarrow \ \boxed{\bold{\frac{39}{13}=3}}$