Question

B= 2² + 2³ + 2⁴ + 2⁵ + ..... + 299 chia hết cho 3

 

Answer 1

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Answer 2

\(2^2+2^3+2^4+2^5+...+2^{99}=2^2\left(1+2\right)+2^4\left(1+2\right)+...+2^{98}\left(1+2\right)=3.2^2+3.2^4+...+3.2^{98}=3\left(2^2+2^4+...+2^{98}\right)⋮3\)

Answer 3

\(B=2^2+2^3+...+2^{99}\)

\(B=\left(2^2+2^3\right)+...+\left(2^4+2^5\right)+...+\left(2^{98}+2^{99}\right)\)

\(B=3.2^2+3.2^4+...+3.2^{98}\)

\(B=3.\left(2^2+2^4+...+2^{98}\right)\)

\(\Rightarrow B⋮3\)

 

Answer 4

Ta có: \(B=2^2+2^3+2^4+2^5+...+2^{98}+2^{99}\)

\(=2^2\left(1+2\right)+2^4\left(1+2\right)+...+2^{98}\left(1+2\right)\)

\(=3\cdot\left(2^2+2^4+...+2^{98}\right)⋮3\)