Question

a=2^0 x 2^1...x2^99x2^100 và chứng minh a chia hết cho 21 

Answer 1

Sửa đề: \(a=2^0+2^1+\cdots+2^{101}\)

Ta có: \(a=2^0+2^1+\cdots+2^{101}\)

\(=\left(1+2+\cdots+2^5\right)+\left(2^6+2^7+\cdots+2^{11}\right)+\cdots+\left(2^{96}+2^{97}+\ldots+2^{101}\right)\)

\(=\left(1+2+\cdots+2^5\right)+2^6\left(1+2+\cdots+2^5\right)+2^{96}\left(1+2+\cdots+2^5\right)\)

\(=63\left(1+2^6+\cdots+2^{96}\right)\) ⋮21