Question

[ke giau mat]

Chứng minh:

2⁰+2¹+2²+...+2^5n-3+2^5n-2+2^{5n-1 chia hết cho 31}

^{(nếu n là số nguyên bất kì)}

Answer 1

[Minh Hiền]

\(\text{Đặt }A=\left(2^0+2^1+2^2+2^3+2^4\right)+...+\left(2^{5n-5}+2^{5n-4}+2^{5n-3}+2^{5n-2}+2^{5n-1}\right)\)

\(=\left(1+2 +4+8+16\right)+...+2^{5n-5}.\left(2^0+2^1+2^2+2^3+2^4\right)\)

\(=31+...+2^{5n-5}.31\)

\(=31.\left(1+...+2^{5n-5}\right)\text{chia hết cho 31}\left(đpcm\right)\)