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

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

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

= \(3^1.4+3^3.4+...+3^{99}.4\)

= \(4.\left(3^1+3^3+...+3^{99}\right)\) chia hết cho 4 

Nên \(3^1+3^2+...+3^{99}+3^{100}\) chia hết cho 4 

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