Tính số dư khi chia: (2^1+2^2+2^3+...+2^99+2^100) cho 7

* Hướng dẫn giải

Ta có

 $2^1$+$2^2+2^3+2^4$+$2^5+2^6+2^7$+...+$2^{98}+2^{99}+2^{100}$

=$2^1$ + ($2^2+2^3+2^4$) + ($2^5+2^6+2^7$) +...+ ($2^{98}+2^{99}+2^{100}$)

$$\begin{gathered} =2+2^2\left(1+2+2^2\right)+2^5\left(1+2+2^2\right)+...+ \\ {2}^{98}\left(1+2+2^2\right) \end{gathered}$$

$$\begin{gathered} =2+2^2.7+2^5.7+...+2^{98}.7 \\ =2+7\left(2^2+2^5+...+2^{98}\right) \end{gathered}$$

Mà $7.\left(2^2+2^5+...+2^{98}\right)⋮7$ 

Nên $2+7\left(2^2+2^5+...+2^{98}\right):7 dư 2$