Question

Chứng minh rằng tổng A chia hết cho 31

A=\(2^0+2^1+2^2+...+2^{2004}\)

Answer 1

\(A=2^0+2^1+2^2+...+2^{2004}\)

\(A=\left(2^0+2^1+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8+2^9\right)+...+\left(2^{2000}+2^{2001}+2^{2002}+2^{2003}+2^{2004}\right)\)

\(A=\left(2^0+2^1+2^2+2^3+2^4\right)+2^5\cdot\left(2^0+2^1+2^2+2^3+2^4\right)+...+2^{2000}\cdot\left(2^0+2^1+2^2+2^3+2^4\right)\)

\(A=31+2^5\cdot31+...+2^{2000}\cdot31\)

\(A=31\cdot\left(1+2^5+...+2^{2000}\right)\)

\(\Rightarrow A⋮31\left(đpcm\right)\)