Đặt A = \(1+2+2^2+2^3+2^4+....+2^{100}\)

2A = \(2\left(1+2+2^2+2^3+2^4+....+2^{100}\right)\)

= \(2+2^2+2^3+2^4+2^5+...+2^{101}\)

2A - A = \(\left(2+2^2+2^3+2^4+2^5+....+2^{101}\right)-\left(1+2^2+2^3+2^4+...+2^{100}\right)\)

= \(2^{101}-1\)

Nếu bạn bt lm r thì ko nên ra câu hỏi nx đâu .

\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

\(\Rightarrow2A=2^2+2^3+2^4+...+2^{100}+2^{101}\)

\(\Rightarrow A=2A-A=2^2+2^3+2^4+...+2^{100}+2^{101}-2-2^2-2^3-2^4-...-2^{99}-2^{100}=2^{101}-2\)

Có : \(S=1+2+2^2+2^3+....+2^{99}\)

\(\Rightarrow2S=2+2^2+2^3+....+2^{100}\)

\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{100}\right)-\left(1+2+2^2+....+2^{99}\right)\)

\(\Rightarrow S=2^{100}-1< 2^{100}\)

Vậy \(S< 2^{100}\)

 S=1+2+2²+2³+....+2⁹⁹

⇒2S=2+2²+2³+....+2¹⁰⁰

⇒2S−S=2¹⁰⁰-1

S=2¹⁰⁰-1

vì 2¹⁰⁰ -1<2¹⁰⁰

⇒S<2¹⁰⁰

\(A=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)

\(=6+2^2.6+...+2^{98}.6=6\left(1+2^2+...+2^{98}\right)⋮6\)

\(A=2+2^2+2^3+...+2^{100}\)

\(=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)

\(=6+6.2^2+...+6.2^{98}\)

\(=6\left(1+2^2+...+2^{98}\right)⋮6\)

\(A=2+2^2+2^3+2^4+...+2^{100}\)

\(=2\cdot3+2^3\cdot3+...+2^{99}\cdot3\)

\(=6\left(1+2^2+...+2^{98}\right)⋮6\)

\(2^{100}-2^{99}+2^{98}-2^{97}+2^{96}-2^{95}+...+2^4-2^3+2^2\)

\(=\left(2^{100}-2^{99}+2^{98}\right)-\left(2^{97}-2^{96}+2^{95}\right)+...+\left(2^4-2^3+2^2\right)\)

\(=2^{96}\left(2^4-2^3+2^2\right)-2^{93}\left(2^4-2^3+2^2\right)+...+\left(2^4-2^3+2^2\right)\)

\(=12\left(2^{96}-2^{93}+...+1\right)⋮12\)

\(A=1+2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

\(\Rightarrow2A=2+2^2+2^3+2^4+2^5+...+2^{100}+2^{101}\)

\(\Rightarrow2A-A=2^{101}-1\)

\(\Leftrightarrow A=2^{101}-1\)

Đặt biểu thức là A

ta có 2A-A=2^101-1

Đặt \(A=1+2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{100}+2^{101}\)

\(\Rightarrow A=2A-A=\left(2+2^2+2^3+2^4+...+2^{101}\right)-\left(1+2+2^2+2^3+...+2^{100}\right)=2^{101}-1\)

Lời giải:
$A=(2+2^2)+(2^3+2^4)+....+(2^{99}+2^{100})$
$=2(1+2)+2^3(1+2)+...+2^{99}(1+2)$

$=2.3+2^3.3+...+2^{99}.3$

$=3(2+2^3+...+2^{99})\vdots 3$

Ta có đpcm.