A=1+2+2²+2³+...+2⁶²+2⁶³

2A=2+2²+2³+2⁴+...+2⁶³+2⁶⁴

2A-A=2+2²+2³+2⁴+...+2⁶³+2⁶⁴-1+2+2²+2³+...+2⁶²+2⁶³

A=2⁶⁴-1

\(A=1+2+2^2+2^3+..+2^{62}+2^{63}\)

\(2A=2+2^2+2^3+...+2^{63}+2^{64}\)

\(2A-A=2^{64}-1\)

\(A=2^{64}-1\)

A=1+2+2²+2³+...+2⁶²+2⁶³

2A=2(1+2+2²+2³+...+2⁶²+2⁶³)

2A=2+2³+2⁴+2⁵...+2⁶³+2⁶⁴

2A-A=(2+2³+2⁴+2⁵...+2⁶³+2⁶⁴)-(1+2+2²+2³+...+2⁶²+2⁶³)

A=2⁶⁴-1

Nha bạn. Chúc bn ht

A = 1 + 2 + 2² + 2³ + ... + 2⁶² + 2⁶³

2A = 2 + 2² + 2³ + 2⁴ + ... + 2⁶³ + 2⁶⁴

A = 2⁶⁴ - 1

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

\(S=1+2+2^2+2^3+2^4+...+2^{2011}\)

\(\Rightarrow S=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{2009}\left(1+2+2^2\right)\)

\(\Rightarrow S=7+2^3.7+...+2^{2009}.7\)

\(\Rightarrow S=7\left(1+2^3+...+2^{2009}\right)⋮7\)

\(\Rightarrow dpcm\)

vì tổng của S chia hết cho 3 nên S chia hết cho 3. có thế cũng hỏi =))

Chúc bạn an toàn

s=[1+2]+[2+2 mũ 2]+...+[2 mũ 6+2 mũ 7]

s=1 nhân [1+2]+2 nhân [1+2]+...+2 mũ 6 nhân [1+2]

s=[1+2] nhân[1+2+...+2 mũ 6

s=3 nhân [1+2+...+2 mũ 6]

=> s chia hết cho 3

\(S=\left(1+2\right)+...+2^6\left(1+2\right)=3\left(1+...+2^6\right)⋮3\)

\(S=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{95}+2^{96}\right)\\ S=\left(1+2\right)\left(2+2^3+...+2^{95}\right)\\ S=3\left(2+2^3+...+2^{95}\right)⋮3\left(1\right)\\ S=\left(2+2^2\right)+2^3\left(1+2^2+...+2^{93}\right)\\ S=8+8\left(1+2^2+...+2^{93}\right)⋮8\left(2\right)\\ \left(1\right)\left(2\right)\Rightarrow S⋮24\)

S=(1+2)+...+2^6(1+2)=3(1+...+2^6)⋮3