Question

Chứng tỏ M=2+2²⁺2³+...+2¹⁰⁰ chia hết cho 31

Answer 1

Ta có:M=(2+2²+2³+2⁴+2⁵)+..........+(2⁹⁶+2⁹⁷+2⁹⁸+2⁹⁹+2¹⁰⁰)

M=2.(1+2+2²+2³+2⁴)+..................+2⁹⁶.(1+2+2²+2³+2⁴)

M=2.31+.............+2⁹⁶.31

M=(2+2⁶+............+2⁹⁶).31 chia hết cho 31(đpcm)

Answer 2

Ta có : \(M=2+2^2+2^3+...+2^{100}\)

=> \(M=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}\right)\)

=> \(M=2\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)

=> \(M=2.31+2^6.31+...+2^{96}.31\)

=> \(M=31\left(2+2^6+...+2^{96}\right)\)

Ta có 31 chia hết cho 31 => M chia hết cho 31

Answer 3

M=2+2²+2³+2⁴+2⁵+...+2¹⁰⁰

M=2(1+2+4+8+16)+...+2⁹⁶(1+2+4+8+16)

M=2.31+...+2⁹⁶.31

M=(2+2⁶+...+2⁹⁶).31

Vì (2+2⁶+...+2⁹⁶).31 chia hết cho 31 nên M chia hết cho 31

Vậy M chia hết cho 31