Question

chung minh rang:S=2+2^2+2^3+....+2^99+2^100 chia het cho 31

Answer 1

Ta có: \(S=2+2^2+2^3+...+2^{99}+2^{100}\)

    \(\Rightarrow S=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{94}+2^{95}+2^{96}+2^{97}+2^{99}\right)\)

\(\Rightarrow S=2.\left(1+2+2^2+2^3+2^4\right)+2^6.\left(1+2+2^2+2^3+2^4\right)+...+2^{94}.\left(1+2+2^2+2^3+2^4\right)\)

\(\Rightarrow2.31+2^6.31+...+2^{94}.31\)

\(\Rightarrow S=31.\left(2+2^6+....+2^{94}\right)\) CHIA HẾT CHO 31 (đpcm)

Vậy S chia hết cho 31