Question

CMR:

\(2+2^2+2^3+2^4+...+2^{60}\) chia hết cho 21

Giúp mik với

 

Answer 1

\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(=7\left(2+2^4+...+2^{55}+2^{58}\right)\)

\(=7\cdot\left[2\left(1+2^3\right)+...+2^{55}\left(1+2^3\right)\right]\)

\(=7\cdot9\cdot\left(2+...+2^{55}\right)⋮21\)

Answer 2

A = 2(1+2+ 2^2)+ 2^4(1+2+2^2)+...+ 2^58(1+2+2^2)=7(2+2^4 +...+2^55+2^58)=7⋅[2(1+2^3)+...+2^55(1+2^3)]=7⋅9⋅(2+...+2^55)⋮21