Question

Chứng tỏ rằng:   3¹ + 3² +3³ + 3⁴ + ... + 3⁹⁹+ 3¹⁰⁰    chia hết cho 4

giúp mình với ạ 

 

Answer 1

Ta có :

\(3^1+3^2+3^3+3^4+...+3^{99}+3^{100}\)

\(=(3^1+3^2)+(3^3+3^4)+...+(3^{99}+3^{100})\)

\(=3(1+3)+3^3(1+3)+...+3^{99}(1+3)\)

\(=3.4+3^3.4+...+3^{99}.4\)

\(=4.(3+3^3+...+3^{99})\)chia hết cho 4 

Answer 2

\(3+3^2+3^3+3^4+...+3^{99}+3^{100}.\)

\(=3\left(1+3\right)+3^2\left(1+3\right)+...+3^{99}\left(1+3\right)\)

\(=4\left(3+3^2+...+3^{99}\right)⋮4\)

Answer 3

Đặt \(A=3^1+3^2+3^3+...+3^{100}\)

\(=\left(3^1+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{99}\left(1+3\right)\)

\(=4\left(3+3^3+...+3^{99}\right)⋮4^{\left(đpcm\right)}\)

Answer 4

\(3^1+3^2+3^3+....+3^{99}+3^{100}\)

\(=\left(1+3\right)+\left(3^2+3^3\right)+.....+\left(3^{99}+3^{100}\right)\)

\(=\left(1+3\right)+3^2\left(1+3\right)+...+3^{99}\left(1+3\right)\)

\(=4+3^2.4+....+3^{99}.4\)

\(=4.\left(1+3^2+....+3^{99}\right)⋮4\)

Answer 5

\(3^1+3^2+3^3+3^4+...+3^{99}+3^{100}\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{99}\left(1+3\right)\)

\(=3.4+3^3.4+...+3^{99}.4⋮4\)