tính tổng s=(-1/70)^0+(-1/7)^1+(-1/7)^2+...+(-1/7)^2016
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Ta có : A= x^0+ x^1+ x^2+...+x^n => \(A=\frac{x^{n+1}-1}{x-1}\)
Chứng minh: xA=x1+x2+...+x^n+1
xA-A=A(x-1)=xn+1-x0=xn+1-1
Từ đó => điều trên
Vậy Ta có:
\(S=\frac{\left(-\frac{1}{7}\right)^{2017}-1}{-\frac{1}{7}-1}\)
Ta có: \(S=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2014}\)
\(\Leftrightarrow\dfrac{-1}{7}\cdot S=\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+\left(-\dfrac{1}{7}\right)^3+...+\left(-\dfrac{1}{7}\right)^{2015}\)
\(\Leftrightarrow S-\dfrac{-1}{7}\cdot S=\left(-\dfrac{1}{7}\right)^0-\left(-\dfrac{1}{7}\right)^{2015}\)
\(\Leftrightarrow\dfrac{8}{7}\cdot S=1+\dfrac{1}{7^{2015}}\)
\(\Leftrightarrow S=\left(1+\dfrac{1}{7^{2015}}\right):\dfrac{8}{7}=\dfrac{\left(1+\dfrac{1}{7^{2015}}\right)\cdot7}{8}\)
S=(\(\dfrac{-1}{7}\))0+(\(\dfrac{-1}{7}\))1+...+(\(\dfrac{-1}{7}\))2016
\(\Rightarrow\)\(\dfrac{-1}{7}S\)=(\(\dfrac{-1}{7}\))1+(\(\dfrac{-1}{7}\))2+...+(\(\dfrac{-1}{7}\))2017
\(\Rightarrow\)\(\dfrac{-1}{7}S\)-\(S\)=\([\) (\(\dfrac{-1}{7}\))1+(\(\dfrac{-1}{7}\))2+...+
(\(\dfrac{-1}{7}\))2017 \(]\)-\([\)(\(\dfrac{-1}{7}\))0+(\(\dfrac{-1}{7}\))1+...+
(\(\dfrac{-1}{7}\))2016\(]\)
=(\(\dfrac{-1}{7}\))1+(\(\dfrac{-1}{7}\))2+...+(\(\dfrac{-1}{7}\))2017-
(\(\dfrac{-1}{7}\))0-(\(\dfrac{-1}{7}\))1-...-(\(\dfrac{-1}{7}\))2016
\(\dfrac{-8}{7}S\)=(\(\dfrac{-1}{7}\))2017-1
S=\(\dfrac{(\dfrac{-1}{7})^{2017}-1}{\dfrac{-8}{7}}\)
S= -(1/7^0 + 1/7^1+ 1/7^2 + 1/7^3 +...+ 1/7^2016)
Xét A = 1/7^0 + 1/7^1 + 1/7^2 + 1/7^3 +...+ 1/7^2016
=>7A= 7 + 1/7^0 + 1/7^1 + ...+ 1/7^2015
=> 6A = 7 - 1/7^2016
=> A = (7 - 1/7^2016)/6
=>S=-(7-1/7^2016)/6
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
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