Question

Cho N=1+3+\(3^2\)+\(3^3\) +........+\(3^{200}\) số N chia hết cho số nào trong các số 2,3,13.

LÀM ƠN ĐI GIÚP MÌNH NHA! MÌNH ĐANG GẤP LẮM LUÔN.

Answer 1

+) Ta có:

\(N=1+3+3^2+3^3+...+3^{200}\)

\(\Rightarrow N=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{199}+3^{200}\right)\)

\(\Rightarrow N=4.3^2\left(1+3\right)+...+3^{199}\left(1+3\right)\)

\(\Rightarrow N=4+3^2.4+...+3^{199}.4\)

\(\Rightarrow N=\left(1+3^2+...+3^{199}\right).4⋮2;⋮̸3\)

\(\Rightarrow N⋮2\) và \(N⋮̸3\)

+) Ta có:

\(N=1+3+3^2+3^3+...+3^{200}\)

\(\Rightarrow N=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{198}+3^{199}+3^{200}\right)\)

\(\Rightarrow N=13+3^3.\left(1+3+3^2\right)+...+3^{198}.\left(1+3+3^2\right)\)

\(\Rightarrow N=13+3^3.13+...+3^{198}.13\)

\(\Rightarrow N=\left(1+3^3+...+3^{198}\right).13⋮13\)

\(\Rightarrow N⋮13\)