Question

tính

\(S=\left(\dfrac{-1}{7}\right)^0+\left(\dfrac{-1}{7}\right)^1+\left(\dfrac{-1}{7}\right)^2+...+\left(\dfrac{-1}{7}\right)^{2016}\)

Answer 1

mk ko chép đề đâu nha

\(S=1+\dfrac{-1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{2016}}\)

đặt \(7S=7-1+\dfrac{1}{7}+...+\dfrac{1}{7^{2015}}\)

=>\(7S+S=\left(7-1+\dfrac{1}{7}+...+\dfrac{1}{7^{2015}}\right)+\left(1-\dfrac{1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{2016}}\right)\)

=>\(8S=7-1+\dfrac{1}{7}+...+\dfrac{1}{7^{2015}}+1-\dfrac{1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{2016}}\)

=>\(8S=7+\left(-1+1\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+...+\left(\dfrac{1}{7^{2015}}-\dfrac{1}{7^{2015}}\right)+\dfrac{1}{7^{2016}}\)

=> \(8S=7+\dfrac{1}{7^{2016}}\)

\(\Rightarrow S=\dfrac{7+\dfrac{1}{7^{2016}}}{8}\)

Answer 2

Gỉa sử : \(-\dfrac{1}{7}=a\)

Thay vào S ,có :

\(a^0+a^1+a^{2^{ }}+.........+a^{2016}\) (1)

=> a.S = a( \(a^0+a^1+a^{2^{ }}+.........+a^{2016}\) )

= \(a^1+a^2+a^3+.........+a^{2016}+a^{2017}\) (2)

Lấy (2) - (1) ,CÓ :

aS-S=( \(a^1+a^2+a^3+.........+a^{2016}+a^{2017}\) ) - ( \(a^0+a^1+a^{2^{ }}+.........+a^{2016}\) ) aS-S= \(a^1+a^2+a^3+.........+a^{2016}+a^{2017}\) - \(1-a-a^2-.........-a^{2016}\)

aS-S = a²⁰¹⁷ -1 => S(a-1) = a²⁰¹⁷ -1

=> S = \(\dfrac{a^{2017}-1}{a-1}\)

Thay a= -1/7 vào S = \(\dfrac{a^{2017}-1}{a-1}\) ,có :

S = \(\dfrac{\left(\dfrac{-1}{7}\right)^{2017}-1}{-\dfrac{1}{7}-1}=\dfrac{\left(-\dfrac{1}{7}\right)^{2017}}{-\dfrac{8}{7}}\)