Question

Tính tổng A=\(\left(-7\right)+\left(-7\right)^2+\left(-7\right)^3+...+\left(-7\right)^{2007}\) Chứng minh rằng A chia hết cho 43

Answer 1

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.

Answer 2

tổng A luôn chia hết nha bạn