Question

Cho \(B=2^1+2^2+2^3+...+2^{30}\). CMR: B chia hết cho 21

Answer 1

Ta có \(B=2^1+2^2+2^3+...+2^{30}\)

\(\Rightarrow2B=2^2+2^3+2^4+...+2^{31}\)

\(\Rightarrow B=2B-B=\)\(\left(2^2+2^3+2^4+...+2^{31}\right)-\left(2^1+2^2+2^3+...+2^{30}\right)\)

\(\Leftrightarrow B=2^{31}-2=2\left(2^{30}-1\right)=2\left(8^{10}-1\right)\)

Mà \(8^{10}-1⋮\left(8-1\right)\Leftrightarrow8^{10}-1⋮7\) (1)

Mặt khác \(8^{10}-1=\left(9-1\right)^{10}-1=BS3+1-1=BS3\left(2\right)\)

(1) ; (2) và (7;3) = 1 \(\Rightarrowđpcm\)