Bài 1:
A = 2 + $2^{2}$ + $2^{3}$ + $2^{4}$ +...+ $2^{100}$
= (2 + $2^{2}$ + $2^{3}$ + $2^{4}$) + ( $2^{5}$ + $2^{6}$ + $2^{7}$ + $2^{8}$)+...+ ($2^{97}$ + $2^{98}$ + $2^{99}$ + $2^{100}$ )
= 30 + $2^{4}$.(2 + $2^{2}$ + $2^{3}$ + $2^{4}$) +...+ $2^{96}$.(2 + $2^{2}$ + $2^{3}$ + $2^{4}$)
=30 + $2^{4}$.30 +...+$2^{96}$.30
= 30.(1+$2^{4}$+...+$2^{96}$)
Vì: 30 chia hết cho 30
=> 30.(1+$2^{4}$+...+$2^{96}$) chia hết cho 30
Nên: A chia hết cho 30
Bài 2:
$4^{x-1}$ + $1^{2019}$ = 257
=> $4^{x-1}$ + 1 = 257
=> $4^{x-1}$ = 257 - 1
=> $4^{x-1}$ = 256
=> $4^{x-1}$ = $4^{4}$
=> x-1 = 4
=> x = 4+1
=> x = 5
Vậy: x = 5
Chúc bạn học tốt!
Mong bạn cho mình 5 sao + cảm ơn + câu trả lời hay nhất ạ!
