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21 tháng 10 2016

Đặt \(A=\left(\frac{-1}{7}\right)^0+\left(\frac{-1}{7}\right)^1+\left(\frac{-1}{7}\right)^2+...+\left(\frac{-1}{7}\right)^{2007}\)

\(\frac{-1}{7}.A=\left(\frac{-1}{7}\right)^1+\left(\frac{-1}{7}\right)^2+\left(\frac{-1}{7}\right)^3+...+\left(\frac{-1}{7}\right)^{2008}\)

\(A-\frac{-1}{7}.A=\left[\left(\frac{-1}{7}\right)^0+\left(\frac{-1}{7}\right)^1+\left(\frac{-1}{7}\right)^2+...+\left(\frac{-1}{7}\right)^{2007}\right]-\left[\left(\frac{-1}{7}\right)^1+\left(\frac{-1}{7}\right)^2+\left(\frac{-1}{7}\right)^3+...+\left(\frac{-1}{7}\right)^{2008}\right]\)

\(A+\frac{1}{7}.A=\left(\frac{-1}{7}\right)^0-\left(\frac{-1}{7}\right)^{2008}\)

\(\frac{8}{7}.A=1-\left(\frac{1}{7}\right)^{2008}\)

\(\frac{8}{7}.A=1-\frac{1}{7^{2008}}\)

\(A=\left(1-\frac{1}{7^{2008}}\right):\frac{8}{7}=\frac{\left(1-\frac{1}{7^{2008}}\right).7}{8}\)

25 tháng 11 2021

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21 tháng 3 2016

S=(-1/7)0+(-1/7)1+...+(-1/7)2007

-1/7.S=(-1/7)1+(-1/7)2+...+(-1/7)2008

-1/7.S-S=[(-1/7)1+(-1/7)2+...+(-1/7)2008]-[(-1/7)0+(-1/7)1+...+(-1/7)2007]

-8/7.S=(-1/7)2008-(-1/7)0

-8/7.S=(1/7)2008-1

.........................

\(A=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+...+\left(-\dfrac{1}{7}\right)^{2007}\)

\(\Leftrightarrow-\dfrac{1}{7}A=\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2008}\)

\(\Leftrightarrow-\dfrac{8}{7}A=\left(-\dfrac{1}{7}\right)^{2008}-1=\dfrac{1}{7^{2008}}-1=\dfrac{1-7^{2008}}{7^{2008}}\)

\(\Leftrightarrow A=\dfrac{1-7^{2008}}{7^{2008}}\cdot\dfrac{-7}{8}=\dfrac{7^{2008}-1}{8\cdot7^{2007}}\)

\(A=\left(-\dfrac{1}{7}\right)^0+\left(\dfrac{-1}{7}\right)^1+...+\left(-\dfrac{1}{7}\right)^{2007}\)

\(\Leftrightarrow\left(-\dfrac{1}{7}\right)A=\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2008}\)

\(\Leftrightarrow-\dfrac{8}{7}A=\left(-\dfrac{1}{7}\right)^{2008}-1\)

\(\Leftrightarrow A=\left(\dfrac{1}{7^{2008}}-\dfrac{7^{2008}}{7^{2008}}\right):\dfrac{-8}{7}=\dfrac{1-7^{2008}}{7^{2008}}\cdot\dfrac{-7}{8}=\dfrac{7^{2008}-1}{8\cdot7^{2007}}\)

11 tháng 12 2015

S=a^0+a^1+a^2+....+a^2007 (1) <=>a.S=a^1+a^2+a^3+....+a^2007+a^2008 (2) lấy (2) trừ (1) ta được: a.S-S=a^2008-a^0=a^2008-1 <=>S=(a^2008-1)/(a-1) với a=-1/7 ta có: S= (-1/7)^0 + (-1/7)^1+(-1/7)^2 +...+ (-1/7)^2007 =[(-1/7)^2008 -1]/(-1/7 -1)

5 tháng 2 2020

a) \(S=\left(-\frac{1}{7}\right)^0+\left(-\frac{1}{7}\right)^1+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2007}\)

\(=1+\left(-\frac{1}{7}\right)+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2007}\)

=> 7S = \(7+\left(-1\right)+\left(-\frac{1}{7}\right)+...+\left(-\frac{1}{7}\right)^{2006}\)

Lấy 7S trừ S ta có : 

7S - S = \(7+\left(-1\right)+\left(-\frac{1}{7}\right)+...+\left(-\frac{1}{7}\right)^{2006}-\left[1+\left(-\frac{1}{7}\right)+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2007}\right]\)

6S = \(7-1-1+\left(\frac{1}{7}\right)^{2007}=5+\left(\frac{1}{7}\right)^{2007}\Rightarrow S=\frac{5+\left(\frac{1}{7}\right)^{2007}}{6}\)