Question

[Nguyễn Duy Khang]

chứng tỏ a chia hết cho 6 với A=2+2 mũ 2 + 2 mũ 3 + 2 mũ 4 + ....+2 mũ 99 + 2 mũ 100

Answer 1

[༺༒༻²ᵏ⁸]

\(A=2+2^2+2^3+...+2^{100}\)

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

\(A=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)

\(A=6+2^2.6+...+2^{98}.6\)

\(A=6\left(1+2^2+...+2^{98}\right)\)

Có : \(6⋮6\)

\(\Rightarrow A=6\left(1+2^2+...+2^{98}\right)⋮6\)

\(\Rightarrow A⋮6\)

Answer 2

[Đinh Lê Tiến Thành]

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