Question

chung minh rang: 1+3+3^2+3^3+...+3^2011 chia het cho 10

Answer 1

Ta có

1+3+3²+3³+...+3²⁰¹¹

= (1+3+3²+3³)+....+(3²⁰⁰⁸+3²⁰⁰⁹+3²⁰¹⁰+3²⁰¹¹)

=40+40+...+40

=10(4+4+...+4)\(⋮\)10 (đpcm)

Answer 2

đặt A= 1+3+3² +........+3²⁰¹¹

=> 3A=3+3² +3³+.......+3²⁰¹¹+3²⁰¹²

=> 3A-A=3²⁰¹²-1

=>A=(3²⁰¹²-1)/2

 

Answer 3

Đặt \(A=1+3+3^2+3^3+.......+3^{2011}\)

\(\Rightarrow A=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+........+\left(3^{2008}+3^{2009}+3^{2010}+3^{2011}\right)\)

\(\Rightarrow A=10+3^4.\left(1+3+3^2+3^3\right)+.......+3^{2008}.\left(1+3+3^2+3^3\right)\)

\(\Rightarrow A=10+3^4.10+.........+3^{2008}.10\)

\(\Rightarrow A=10\left(1+3^4+......+3^{2008}\right)⋮10\)( đpcm )

Vậy .....