Question

Tổng sau có chia hết cho 3 ko:

\(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^{10}\)

Answer 1

Ta có :

\(2+2^2+2^3+....+2^{10}\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+....+2^9\left(1+2\right)\)

\(=2.3+2^3.3+....+2^9.3\)

=> Tổng chia hết cho 3

Answer 2

\(2+2^2+2^3+...+2^{10}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)\)

\(=2.3+2^3.3+...+2^9.3\)

\(=\left(2+2^3+...+2^9\right).3⋮3\)

\(\Rightarrow2+2^2+2^3+...+2^{10}⋮3\)