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: \(S=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2014}\)

\(\Leftrightarrow\dfrac{-1}{7}\cdot S=\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+\left(-\dfrac{1}{7}\right)^3+...+\left(-\dfrac{1}{7}\right)^{2015}\)

\(\Leftrightarrow S-\dfrac{-1}{7}\cdot S=\left(-\dfrac{1}{7}\right)^0-\left(-\dfrac{1}{7}\right)^{2015}\)

\(\Leftrightarrow\dfrac{8}{7}\cdot S=1+\dfrac{1}{7^{2015}}\)

\(\Leftrightarrow S=\left(1+\dfrac{1}{7^{2015}}\right):\dfrac{8}{7}=\dfrac{\left(1+\dfrac{1}{7^{2015}}\right)\cdot7}{8}\)

Cười cái gì mà cười

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

7s-s

tich cho mk nha

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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