Question

Bài 4:Tính tổng A = (-7) + (- 7)2 + (- 7)3 + ....+( -7)2007 .CMR A chia hết cho 43

Answer 1

Ta có:

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

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

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

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

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

\(\Rightarrow A⋮43\)

Vậy \(A⋮43\)

Answer 2

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

(-7).A=(-7^2)+(-7^3)+...(-7^2007)+(-7^2008)

=>A-(-7)A=(-7)-(-7^2008)

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

b) A={(-7)+(-7^20)+(-7^3)}+...+{(-7^2005)+(-7^2006)+(-7^2007) (chia thành 2007:3=669 nhóm 3 số)

A=(-7).{1+(-7)+(-7^2)}+...+(-2007^2005).{1+(-7)+(-7^2)}

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

Vậy A chia hết cho 43