Question

2+2^2+2^3+2^4+...+2^100+2^101.Tìm số dư của A khi chia cho 7

Answer 1

Lời giải:

$A=2+2^2+2^3+2^4+...+2^{100}+2^{101}$

$=2+2^2+(2^3+2^4+2^5)+....+(2^{99}+2^{100}+2^{101})$

$=6+2^3(1+2+2^2)+....+2^{99}(1+2+2^2)$

$=6+(1+2+2^2)(2^3+....+2^{99})$

$=6+7(2^3+....+2^{99})$

$\Rightarrow A$ chia $7$ dư $6$.