Giải:
\(3^1+3^2+3^3+3^4+...+3^{99}+3^{100}\)
\(=\left(3^1+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{99}\left(1+3\right)\)
\(=3.4+3^3.4+...+3^{99}.4\)
\(=4\left(3+3^3+...+3^{99}\right)⋮4\)
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