Question

cho A =1+2+2²+2³+...+2⁹⁹

a.chứng minh rằng A.chia hết cho 3 và 15

Answer 1

\(A=1+2+2^2+...+2^{99}\)

\(A=\left(1+2+2^2+2^3\right)+\left(2^4+2^5+2^6+2^7\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}\right)\)\(A=15+2^4\left(1+2+2^2+2^3\right)+...+2^{96}\left(1+2+2^2+2^3\right)\)

\(A=1.15+2^4.15+...+2^{96}.15\)

\(A=3.5\left(1+2^4+...+2^{96}\right)\)chia hết cho 3 và 15