Question

Cho \(A=2^1+2^2+2^3+...+2^{120}\)Chứng minh rằng A chia hết cho 21

Answer 1

Ta có:

A=2+2²+2³+...+2¹²⁰

A=(2+2²+2³+2⁴+2⁵)+...+(2¹¹⁶+2¹¹⁷+2¹¹⁸+2¹¹⁹+2¹²⁰)

A=2.(1+2+2²+2³+2⁴)+...+2¹¹⁶.(1+2+2²+2³+2⁴)

A=2.63+...+2¹¹⁶.63

A=63.(2+...+2¹¹⁶)

A=21.3.(2+...+2¹¹⁶)\(⋮\)21

Vậy A chia hết cho 21

Answer 2

\(A=2^1+2^2+2^3+2^4+....+2^{119}+2^{120}\)

\(=\left(2^1+2^2+2^3+2^4+2^5+2^6\right)+.....+\left(2^{115}+2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=2\left(1+2+2^2+2^3+2^4+2^5\right)+.....+2^{115}\left(1+2+2^2+2^3+2^4+2^5\right)\)

\(=2.63+....+2^{115}.63\)

\(=63\left(2+....+2^{115}\right)\)

\(=3.21.\left(2+...+2^{115}\right)\)

\(\Rightarrow A⋮21\)