Question

Cho: S=2+2²+2³+...+2²⁰⁰⁰

a)Hỏi S có chia hết cho 6 không?

b) Thu gọn S

Answer 1

a ) \(S=2+2^2+2^3+2^4+...+2^{1999}+2^{2000}\)

\(=\left(2+2^2\right)+\left(2^2.2+2^2.2^2\right)+...+\left(2^{1998}.2+2^{1998}.2^2\right)\)

\(=\left(2+4\right)+2^2.\left(2+2^2\right)+..+2^{1998}.\left(2+2^2\right)\)

\(=6+2^6.6+...+2^{1998}.6\)

\(=6.\left(1+2^2+...2^{1998}\right)⋮6\)

\(\Rightarrow S⋮6\)

b ) \(S=2+2^2+2^3+...+2^{2000}\)

 \(\Rightarrow2S=2.\left(2+2^2+2^3+...+2^{2000}\right)\)

\(\Rightarrow2S=2^2+2^3+2^4+...+2^{2001}\)

\(\Rightarrow2S-S=\left(2^2+2^3+2^4+...+2^{2001}\right)-\left(2+2^2+2^3+...+2^{2000}\right)\)

\(\Rightarrow S=2^{2001}-2\)