.Chứng minh rằng
A = (-7)1 + (-7)2 + ....+ (-7)2007 Chia hết cho 43
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\(A=\left(-7\right)+\left(-7\right)^2+\left(-7\right)^3+\left(-7\right)^4+\left(-7\right)^5+\left(-7\right)^6+...+\left(-7\right)^{2005}+\left(-7\right)^{2006}+\left(-7\right)^{2007}\)
\(A=\left[\left(-7\right)+\left(-7\right)^2+\left(-7\right)^3\right]+\left[\left(-7\right)^4+\left(-7\right)^5+\left(-7\right)^6\right]+...+\left[\left(-7\right)^{2005}+\left(-7\right)^{2006}+\left(-7\right)^{2007}\right]\)
\(A=\left(-7\right)\left(1+-7+7^2\right)+\left(-7\right)^4\left(1+-7+7^2\right)+...+\left(-7\right)^{2005}\left(1+-7+7^2\right)\)
\(A=\left(-7\right)\cdot43+\left(-7\right)^4\cdot43+...+\left(-7\right)^{2005}\cdot43\)
\(A=43\left[\left(-7\right)+\left(-7\right)^4+...+\left(-7\right)^{2008}\right]⋮43\left(đpcm\right)\)
Ta có : A = -7 + (-7)2 + (-7)3 + ....... + (-7)2007
=> -7A = (-7)2 + (-7)3 + ....... + (-7)2008
=> -7A - A = (-7)2008 - (-7)
=> -8A = (-7)2008 + 7
=> A = .........................
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.
\(A=\left(-7\right)+\left(-7\right)^2+......+\left(-7\right)^{2006}+\left(-7\right)^{2007}\)
\(=\left[\left(-7\right)+\left(-7\right)^2+\left(-7\right)^3\right]+\left[\left(-7\right)^4+\left(-7\right)^5+\left(-7\right)^6\right]+.......\) \(+\left[\left(-7\right)^{2005}+\left(-7\right)^{2006}+\left(-7\right)^{2007}\right]\)
\(=\left(-7\right)\left[1+\left(-7\right)+\left(-7\right)^2\right]+......+\left(-7\right)^{2005}\left[1+\left(-7\right)+\left(-7\right)^2\right]\)
\(=\left(-7\right).43+\left(-7\right)^3.43+......+\left(-7\right)^{2005}.43\)
\(=43\left[\left(-7\right)+\left(-7\right)^3+.....+\left(-7\right)^{2005}\right]\).
Suy ra A chia hết cho 43.
A=(-7+-7^2+-7^3)+.....+(-7^2005+-7^2006+-7^2007)
A=-7(1+-7+-7^2)+.....+-7^2005(1+-7+-7^2)
A=-7.43+....+-7^2005.43\(⋮\)43\(\Rightarrow\)dpcm
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)
TA CÓ: A= (( -7 ) + ( -7 )2+ ( -7 )3)+...+ ( -7 )2007
A= -7( 1+ ( -7 )+ ( -7 )2) +...+ (-7)2005( 1+ ( -7 ) + ( -7 )2)
A= -7.43+...+ (-7)2005.43
A= 43[ -7+...+ ( -7 )2005 ] CHIA HẾT CHO 43
VẬY A CHIA HẾT CHO 43