Question

CHO A=1+2+2²+......2⁹⁰

^{CMR A⋮7}

Answer 1

\(A=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{98}\left(1+2+2^2\right)\)

\(A=\left(1+2+2^2\right)\left(1+2^3+...+2^{98}\right)\)

A=7(1+2³+...+2⁹⁸)

=>A \(⋮\)7

Answer 2

A=1+2+2²+2³+....+2⁹⁰

A=(1+2+2²)+....+2⁸⁸(1+2+2²)

A=(1+2+2²)(1+2³+...+2⁸⁸)

A=7.(1+2³+...+2⁸⁸)\(⋮\)7

=>A\(⋮\)7