Question

CMR; 3^2+3^3+...+3^101 chia hết cho 120

Answer 1

Đặt \(A=3^2+3^3+3^4+...+3^{101}\)

\(A=\left(3^2+3^3+3^4+3^5\right)+\left(3^6+3^7+3^8+3^9\right)+...+\left(3^{98}+3^{99}+3^{100}+3^{101}\right)\)

\(A=3\left(3+3^2+3^3+3^4\right)+3^5\left(3+3^2+3^3+3^4\right)+...+3^{97}\left(3+3^2+3^3+3^4\right)\)

\(A=3.\left(3+9+27+81\right)+3^5\left(3+9+27+81\right)+...+3^{97}\left(3+9+27+81\right)\)

\(A=3.120+3^5.120+...+3^{97}.120\)

\(A=120\left(3+3^5+...+3^{97}\right)⋮120\)

Vậy \(A⋮120\)

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