Tính :
B = \(\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+....+\left(-\dfrac{1}{7}\right)^{2018}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NT
24 tháng 5 2017
\(S=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+....+\left(-\dfrac{1}{7}\right)^{2017}\\ =1+-\dfrac{1}{7}+\dfrac{1}{7^2}+-\dfrac{1}{7^3}+.....+-\dfrac{1}{7^{2017}}\\ =\left(1+\dfrac{1}{7^2}+\dfrac{1}{7^4}+...+\dfrac{1}{7^{2016}}\right)-\left(\dfrac{1}{7}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{2017}}\right)\)
rồi bạn tính 2 về rồi trừ ra là xng nhé
MV
12 tháng 8 2017
\(S=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2017}\\ S=\dfrac{\left(-1\right)^0}{7^0}+\dfrac{\left(-1\right)^1}{7^1}+\dfrac{\left(-1\right)^2}{7^2}+...+\dfrac{\left(-1\right)^{2017}}{7^{2017}}\\ S=\dfrac{1}{7^0}+\dfrac{-1}{7^1}+\dfrac{1}{7^2}+...+\dfrac{-1}{7^{2017}}\\ -7S=\dfrac{-7}{7^0}+\dfrac{7}{7^1}+\dfrac{-7}{7^2}+...+\dfrac{7}{7^{2017}}\\ -7S=\left(-7\right)+\dfrac{1}{7^0}+\dfrac{-1}{7^1}+...+\dfrac{1}{7^{2016}}\\ -7S-S=\left[\left(-7\right)+\dfrac{1}{7^0}+\dfrac{-1}{7^1}+...+\dfrac{1}{7^{2016}}\right]+\left(\dfrac{1}{7^0}+\dfrac{-1}{7^1}+\dfrac{1}{7^2}+...+\dfrac{-1}{7^{2017}}\right)\\ -8S=\left(-7\right)+\dfrac{-1}{2017}\\ -8S=-\left(7+\dfrac{1}{2017}\right)\\ 8S=7+\dfrac{1}{2017}\\ S=\dfrac{7+\dfrac{1}{2017}}{8}\)
Vậy ...
16 tháng 6 2022
Bài 1:
a: \(A=\left(-\dfrac{1}{5}\right)^{33}:\left(-\dfrac{1}{5}\right)^{32}=\dfrac{-1}{5}\)
c: \(C=\dfrac{2^{12}\cdot3^{10}+3^9\cdot2^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
AN
25 tháng 12 2017
S= \(\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2017}\)
\(\left(-\dfrac{1}{7}\right)S=\left(-\dfrac{1}{7}\right)\left(-\dfrac{1}{7}+-\dfrac{1^2}{7}+..+-\dfrac{1^{2007}}{7}\right)\)
= \(-\dfrac{1}{7}+-\dfrac{1}{7}^2+....+\dfrac{-1^{2008}}{7}\)
=>\(-\dfrac{1}{7}S-S=\) \(-\dfrac{1}{7}+-\dfrac{1}{7}^2+....+\dfrac{-1^{2008}}{7}\) \(-\)\(\left(1+-\dfrac{1}{7}+-\dfrac{1^2}{7}+...+-\dfrac{1^{2007}}{7}\right)\)
=> \(-\dfrac{1}{7}S=\) \(\dfrac{-1^{2008}}{7}-1\)
=> S= \(\dfrac{-1^{2008}}{7}-1\) : \(\dfrac{-1}{7}\)
30 tháng 11 2017
Các bạn trả lời giúp mk nha. Mk đang cần gấp. Chều nay mk kiểm tra rồi
9 tháng 5 2017
mk ko chép đề đâu nha
\(S=1+\dfrac{-1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{2016}}\)
đặt \(7S=7-1+\dfrac{1}{7}+...+\dfrac{1}{7^{2015}}\)
=>\(7S+S=\left(7-1+\dfrac{1}{7}+...+\dfrac{1}{7^{2015}}\right)+\left(1-\dfrac{1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{2016}}\right)\)
=>\(8S=7-1+\dfrac{1}{7}+...+\dfrac{1}{7^{2015}}+1-\dfrac{1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{2016}}\)
=>\(8S=7+\left(-1+1\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+...+\left(\dfrac{1}{7^{2015}}-\dfrac{1}{7^{2015}}\right)+\dfrac{1}{7^{2016}}\)
=> \(8S=7+\dfrac{1}{7^{2016}}\)
\(\Rightarrow S=\dfrac{7+\dfrac{1}{7^{2016}}}{8}\)
9 tháng 5 2017
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 = \(\dfrac{\left(\dfrac{-1}{7}\right)^{2017}-1}{-\dfrac{1}{7}-1}=\dfrac{\left(-\dfrac{1}{7}\right)^{2017}}{-\dfrac{8}{7}}\)
\(B=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2018}\)
\(\Rightarrow-\dfrac{1}{7}B=\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+\left(-\dfrac{1}{7}\right)^3+...+\left(-\dfrac{1}{7}\right)^{2019}\)
\(\Rightarrow-\dfrac{1}{7}B-1=\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+\left(-\dfrac{1}{7}\right)^3+...+\left(-\dfrac{1}{7}\right)^{2019}-\left(-\dfrac{1}{7}\right)^0-\left(-\dfrac{1}{7}\right)^1-\left(-\dfrac{1}{7}\right)^2-...-\left(-\dfrac{1}{7}\right)^{2018}\)
\(\Rightarrow-\dfrac{8}{7}B=\left(-\dfrac{1}{7}\right)^{2019}-1\)
\(\Rightarrow B=\left[\left(-\dfrac{1}{7}\right)^{2019}-1\right]:\left(-\dfrac{8}{7}\right)\)
\(B=1-\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}+...-\dfrac{1}{7^{2017}}+\dfrac{1}{7^{2018}}\\ \Rightarrow7B=7-1+\dfrac{1}{7}-\dfrac{1}{7^2}+...-\dfrac{1}{7^{2016}}+\dfrac{1}{7^{2017}}\\ \Rightarrow7B+B=6+\dfrac{1}{7}-\dfrac{1}{7^2}+...+\dfrac{1}{7^{2017}}+1-\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}+...-\dfrac{1}{7^{2017}}+\dfrac{1}{7^{2018}}\\ \Rightarrow8B=7+\dfrac{1}{7^{2018}}=\dfrac{7^{2019}+1}{7^{2018}}\\ \Rightarrow B=\dfrac{7^{2019}+1}{8\cdot7^{2018}}\)