Question

A= 1+3+3^2+...+3^2006

1) tính 3A

2) chứng minh A=3^2007-1/2

Answer 1

`1,` Ta có:

`A = 1 + 3 + 3^2 + ... + 3^2006`

`3A = 3 . (1 + 3 + 3^2 + ... + 3^2006)`

`3A=  3+ 3^2 + 3^3+ ... + 3^2007`

`2,` Ta có:

`A = 1 + 3 + 3^2 + ... + 3^2006`

`3A = 3 . (1 + 3 + 3^2 + ... + 3^2006)`

`3A=  3+ 3^2 + 3^3+ ... + 3^2007`

`3A - A = (3 + 3^2 + 3^3 + ... + 3^2007) - (1 + 3 + 3^2 + ... + 3^2006)`

`2A = 3^2007 - 1`

`-> A = (3^2007 - 1)/2 (đpcm)`

 

Answer 2

1) \(A=1+3+3^2+...+3^{2006}\)

\(3A=3.\left(1+3+3^2+...+3^{2006}\right)\)

\(3A=3+3^2+3^3+...+3^{2007}\)

2) \(3A-A=\left(3+3^2+3^3+...+3^{2007}\right)-\left(1+3+3^2+...+3^{2006}\right)\)

\(\left(3-1\right)A=3+3^2+3^3+...+3^{2006}-1-3-3^2+...+3^{2006}\)

\(2A=3^{2007}-1\)

\(A=\dfrac{3^{2007}-1}{2}\)

 

Answer 3

\(1,A=1+3+3^2+...+3^{2006}\)

\(3A=3+3^2+3^3+...+3^{2007}\)

\(2,A=1+3+3^2+...+3^{2006}\)

\(3A=3+3^2+3^3+...+3^{2007}\)

\(3A-A=3+3^2+3^3+...+3^{2007}-\left(1+3+3^2+...+3^{2006}\right)\)

\(2A=3^{2007}-1\)

\(A=\dfrac{3^{2007}-1}{2}\)