Question

A = 2 + 2\(^2\) + 2\(^3\) + ... + 2\(^{24}\) 

a. Chứng minh A chia hết cho 14 ; 15

b, Chứng minh A chia hết cho 16 dư 14 

Answer 1

a) Ta có:
\(A=2+2^2+2^3+...+2^{24}\)

\(\Rightarrow A=\left(2+2^2+2^3\right)+...+\left(2^{22}+2^{23}+2^{24}\right)\)

\(\Rightarrow A=14+...+2^{21}.\left(2+2^2+2^3\right)\)

\(\Rightarrow A=14+...+2^{21}.14\)

\(\Rightarrow A=\left(1+...+2^{21}\right).14⋮14\)( đpcm )

\(A=2+2^2+2^3+...+2^{24}\)

\(\Rightarrow A=\left(2+2^2+2^3+2^4\right)+...+\left(2^{21}+2^{22}+2^{23}+2^{24}\right)\)

\(\Rightarrow A=2\left(1+2+2^2+2^3\right)+...+2^{21}\left(1+2+2^2+2^3\right)\)

\(\Rightarrow A=2.15+...+2^{21}.15\)

\(\Rightarrow A=15\left(2+...+2^{21}\right)⋮15\left(đpcm\right)\)

 

 

Answer 2

b) Mk sửa đề chút là A chia 16 dư 15 nhé

Ta có:

\(A=2+2^2+2^3+...+2^{24}\)

\(\Rightarrow A=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{20}+2^{21}+2^{22}+2^{23}+2^{24}\right)\)

\(\Rightarrow A=2\left(1+2+2^2+2^3+2^4\right)+...+2^{20}\left(1+2+2^2+2^3+2^4\right)\)

\(\Rightarrow A=2.31+...+2^{20}.31\)

\(\Rightarrow A=\left(2+2^{20}\right).31\) 

Vì 31 chia 16 dư 15 nên suy ra đpcm