Bổ xung cho chuyên gia

Cách 2

 `a) A=2^1+2^2+2^3+2^4+...+2^{60}`

`⇒A=(2^1+2^2)+(2^3+2^4)+...+(2^{59}+2^{60})`

`⇒A=2×(1+2)+2^3×(1+2)+....+2^{59}×(1+2)`

`⇒A=2×3+2^3×3+...+2^{59}×3`

`⇒A=3×(2+2^3+...+2^{59})`

`⇒A\vdots 3`(đpcm)

`b) A=2^1+2^2+2^3+...+2^{60}`

`⇒A=(2^1+2^2+2^3)+...+(2^{58}+2^{59}+2^{60})`

`⇒A=2×(1+2+2^2)+....+2^{58}×(1+2+2^2)`

`⇒A=2×7+...+2^{58}×7`

`⇒A=7×(2+...+2^{58})`

`⇒A \vdots 7`(đpcm)

`c) A=2^1+2^2+2^3+2^4+...+2^{60}`

`⇒A=(2^1+2^3+2^4)+...+(2^{57}+2^{59}+2^{60})`

`⇒A=2×(1+2^2+2^3)+...+2^{57}×(1+2^2+2^3)`

`⇒A=2×13+..+2^{57}×13`

`⇒A=13×(2+...+2^{57}`

`⇒A \vdots 13` (đpcm)

Giải thích các bước giải:

 Ta có:

$\begin{array}{l}
A = {2^1} + {2^2} + {2^3} + {2^4} + ... + {2^{59}} + {2^{60}}\\
 \Rightarrow 2A = {2^2} + {2^3} + {2^4} + {2^5} + ... + {2^{60}} + {2^{61}}\\
 \Rightarrow 2A - A = \left( {{2^2} + {2^3} + {2^4} + {2^5} + ... + {2^{60}} + {2^{61}}} \right) - \left( {{2^1} + {2^2} + {2^3} + {2^4} + ... + {2^{59}} + {2^{60}}} \right)\\
 \Rightarrow A = {2^{61}} - {2^1}\\
 \Rightarrow A = {2^{61}} - 2
\end{array}$

+) Chứng minh: $A\vdots 3$

Ta có:

$\begin{array}{l}
4 \equiv 1\left( {\bmod 3} \right)\\
 \Leftrightarrow {2^2} \equiv 1\left( {\bmod 3} \right)\\
 \Rightarrow {\left( {{2^2}} \right)^{30}} \equiv 1\left( {\bmod 3} \right)\\
 \Rightarrow {2^{60}} \equiv 1\left( {\bmod 3} \right)\\
 \Rightarrow {2^{61}} \equiv 2\left( {\bmod 3} \right)\\
 \Rightarrow {2^{61}} - 2 \equiv 0\left( {\bmod 3} \right)\\
 \Rightarrow A \equiv 0\left( {\bmod 3} \right)\\
 \Rightarrow A \vdots 3
\end{array}$

+) Chứng minh: $A\vdots 7$

$\begin{array}{l}
64 \equiv 1\left( {\bmod 7} \right)\\
 \Leftrightarrow {2^6} \equiv 1\left( {\bmod 7} \right)\\
 \Rightarrow {\left( {{2^6}} \right)^{10}} \equiv 1\left( {\bmod 7} \right)\\
 \Rightarrow {2^{60}} \equiv 1\left( {\bmod 7} \right)\\
 \Rightarrow {2^{61}} \equiv 2\left( {\bmod 7} \right)\\
 \Rightarrow {2^{61}} - 2 \equiv 0\left( {\bmod 7} \right)\\
 \Rightarrow A \equiv 0\left( {\bmod 7} \right)\\
 \Rightarrow A \vdots 7
\end{array}$

+) Chứng minh: $A\vdots 13$

$\begin{array}{l}
64 \equiv  - 1\left( {\bmod 13} \right)\\
 \Leftrightarrow {2^6} \equiv  - 1\left( {\bmod 13} \right)\\
 \Rightarrow {\left( {{2^6}} \right)^{10}} \equiv {\left( { - 1} \right)^{10}} = 1\left( {\bmod 13} \right)\\
 \Rightarrow {2^{60}} \equiv 1\left( {\bmod 13} \right)\\
 \Rightarrow {2^{61}} \equiv 2\left( {\bmod 13} \right)\\
 \Rightarrow {2^{61}} - 2 \equiv 0\left( {\bmod 13} \right)\\
 \Rightarrow A \equiv 0\left( {\bmod 13} \right)\\
 \Rightarrow A \vdots 13
\end{array}$

Ta có điều phải chứng minh.