Ta thấy \(A=\left(-7\right)+\left(-7\right)^2+\left(-7\right)^3+...+\left(-7\right)^{2007}\)

\(A=\left[\left(-7\right)+\left(-7\right)^2+\left(-7\right)^3\right]+...+\left[\left(-7\right)^{2005}+\left(-7\right)^{2006}+\left(-7\right)^{2007}\right]\)

\(A=-7.\left[1+\left(-7\right)+49\right]+\left(-7\right)^4.\left[1+\left(-7\right)+49\right]+...+\left(-7\right)^{2005}.\left[1+\left(-7\right)+49\right]\)

\(A=-7.43+\left(-7\right)^4.43+...+\left(-7\right)^{2005}.43\)

\(A=43\left[\left(-7\right)+\left(-7\right)^4+...+\left(-7\right)^{2005}\right]⋮43\)

Vậy A chia hết cho 43.

tổng A luôn chia hết nha bạn

\(A=\left(-7\right)+\left(-7\right)^2+...+\left(-7\right)^{2006}+\left(-7\right)^{2007}\)

\(\left(-7\right).A=\left(-7\right)^2+\left(-7\right)^3+...+\left(-7\right)^{2007}+\left(-7\right)^{2008}\)

=> \(A-\left(-7\right)A=\left(-7\right)-\left(-7\right)^{2008}\)

=> \(8A=-7-7^{2008}\) => \(A=-\frac{7+7^{2008}}{8}\)

b) \(A=\left(\left(-7\right)+\left(-7\right)^2+\left(-7\right)^3\right)+...+\left(\left(-7\right)^{2005}+\left(-7\right)^{2006}+\left(-7\right)^{2007}\right)\) ( Chia thành 2007 : 3 = 669 nhóm 3 số)

 \(A=\left(-7\right).\left(1+\left(-7\right)+\left(-7\right)^2\right)+...+\left(-7\right)^{2005}.\left(1+\left(-7\right)+\left(-7\right)^2\right)\)

\(A=\left(-7\right).43+...+\left(-7\right)^{2005}.43=43.\left(\left(-7\right)+...+\left(-7\right)^{2005}\right)\)chia hết cho 43

Vậy A chia hết cho 43

A= (- 7) + (-7)^2+ … + (- 7)^2006 + (- 7)^2007 

<=> -7A = (-7)^2+ … + (- 7)^2006 + (- 7)^2008 

A-(- 7A )= (- 7) + (-7)^2+ … + (- 7)^2006 + (- 7)^2007-{(-7)^2+ … + (- 7)^2006 + (- 7)^2008} 

<=> 8A = -7 - (- 7)^2008 = -7 + 7^2008 = 7^2008 - 7 

<=> A = (7^2008 - 7)/8 .

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Sửa đề: Tính tổng:

\(A=\left(-7\right)+\left(-7\right)^2+...+\left(-7\right)^{2007}...\)

Giải:

\(A=\left(-7\right)+\left(-7\right)^2+...+\left(-7\right)^{2007}\)

\(\Rightarrow-7A=-7\)\(\left[\left(-7\right)+\left(-7\right)^2+...+\left(-7\right)^{2007}\right]\)

\(=\left(-7\right)^2+\left(-7\right)^3+...+\left(-7\right)^{2008}\)

\(\Rightarrow A-\left(-7\right)A=\left(-7\right)-\left(-7\right)^{2008}\)

\(\Rightarrow8A=-7+7^{2008}\Rightarrow A=\dfrac{-7+7^{2008}}{8}\)

Vậy \(A=\dfrac{-7+7^{2008}}{8}\)

_____________________________________

Ta có:

\(A=\left(-7\right)+\left(-7\right)^2+...+\left(-7\right)^{2007}\)

\(=\left[\left(-7\right)+\left(-7\right)^2+\left(-7\right)^3\right]+...+\left[\left(-7\right)^{2005}+\left(-7\right)^{2006}+\left(-7\right)^{2007}\right]\)

\(=\left(-7\right).\left[1+\left(-7\right)+\left(-7\right)^2\right]+...+\left(-7\right)^{2005}\left[1+\left(-7\right)+\left(-7\right)^2\right]\)

\(=\left(-7\right).43+...+\left(-7\right)^{2005}.43\)

\(=43.\left[\left(-7\right)+...+\left(-7\right)^{2005}\right]⋮43\) (Đpcm)

đề sai con cuối

A = 1 . (-7) + (-7) . (-7) + (-7) . \(^{\left(-7\right)^2}\)\(+....+1.\left(-7\right)^{2005}+\left(-7\right).\left(-7\right)^{2005}+\left(-7\right)^2.\left(-7\right)^{2005}\)

\(A=\left(-7\right).\left(1+\left(-7\right)+\left(-7\right)^2\right)+...+\left(-7\right)^{2005}.\left(1+\left(-7\right)+\left(-7\right)^2\right)\)

\(A=\left(-7\right).43+....+\left(-7\right)^{2005}.43\)

\(A=43.\left(\left(-7\right)+.....+\left(-7\right)^{2005}\right)\)chia hết cho 43

Vậy A chia hết cho 43

sao tự nhiên lại có dấu = trong [=7] thế kia

Khó thế!

\(A=\left(-7\right)+\left(-7\right)^2+....+\left(-7\right)^{2007}\)

\(A=-\left(7+7^2+...+7^{2007}\right)\)

\(7A=-\left(7^2+7^3+....+7^{2008}\right)\)

7A-A=6A= 7²⁰⁰⁸- 7

=> A= \(\frac{7^{2008}-7}{6}\)

Mình làm vậy ko biết có đúng ko nữa

Nguyễn Huy Thắng, Nguyễn Huy Tú