K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
DT
0
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
.........................
TH
1
CC
19 tháng 1 2020
\(S=\left(\frac{-1}{7}\right)^0+\left(\frac{-1}{7}\right)^1+...........+\left(\frac{-1}{7}\right)^{2007}\)
\(\Rightarrow\frac{-1}{7}S=\left(\frac{-1}{7}\right)^1+\left(\frac{-1}{7}\right)^2+........+\left(\frac{-1}{7}\right)^{2008}\)
\(\Rightarrow\frac{-1}{7}S-S=\frac{-8}{7}S=\left(\frac{-1}{7}\right)^{2008}-\left(\frac{-1}{7}\right)^0\)
\(\Rightarrow S=\frac{\left(\frac{-1}{7}\right)^{2008}-1}{\frac{-8}{7}}\)
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 ...
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= \(\dfrac{-1^{2008}}{7}-1\) : \(\dfrac{-1}{7}\)
NT
3 tháng 1 2017
S=1+(-1/7)^1+(-1/7)^2+...+(-1/7)^2007
=>7S=7+(-1/7)^1+(1/7)^2+...+(-1/7)^2006
=>(7-1)S=6-(1/7)^2007
=>S=1-(-1/7^2007/6)
3 tháng 1 2017
1/7S=(-1/7)^1+...+(-1/7)2018
1/7S-S=(-1/7)^1+....+(-1/7)^2018-(-1/7)^0-...-(-1/7)^2017
-6/7S=(-1/7)^2018-1=(-1/7)^2018-1:-6/7
NI
17 tháng 1 2017
S=(−1/7)^0+(−1/7)^1+(−1/7)^2+...+(−1/7)^2007
7S = 1+(−1/7)^1+(−1/7)^2+...+(−1/7)^2007
=> 7S = 7+(−1/7)^1+(−1/7)^2+...+(−1/7)^2006
=> 6S = 6-(−1/7)^2007
=> S= 1-(−1/7^2007/6)
23 tháng 6 2022
\(A=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+...+\left(-\dfrac{1}{7}\right)^{2007}\)
\(\Leftrightarrow-\dfrac{1}{7}A=\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2008}\)
\(\Leftrightarrow-\dfrac{8}{7}A=\left(-\dfrac{1}{7}\right)^{2008}-1=\dfrac{1}{7^{2008}}-1=\dfrac{1-7^{2008}}{7^{2008}}\)
\(\Leftrightarrow A=\dfrac{1-7^{2008}}{7^{2008}}\cdot\dfrac{-7}{8}=\dfrac{7^{2008}-1}{8\cdot7^{2007}}\)
Answer:
`S=(-1/7)^0+(-1/7)^1+(-1/7)^2+...+(-1/7)^2007`
`=>S=1-1/7+(1/7)^2-...-(1/7)^2007`
`=>7S=7-1+1/7-...-(1/7)^2006`
`=>S+7S=(1-1/7+(1/7)^2-...-(1/7)^2007)+(7-1+1/7-...-(1/7)^2006)`
`=>8S=7-(1/7)^2017`
`=>8S=7-\frac{1}{7^2007}`
`=>8S=\frac{7^2008-1}{7^2007}`
`=>S=\frac{7^2008-1}{8.7^2007}`