Tính S=(-1/7)^0+(-1/7)^1+(-1/7)^2+....+(-1/7)^2016
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NH
1
7 tháng 5 2016
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:
\(S=\frac{\left(-\frac{1}{7}\right)^{2017}-1}{-\frac{1}{7}-1}\)
AP
0
PN
6 tháng 5 2016
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
TN
0
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 = \(\dfrac{\left(\dfrac{-1}{7}\right)^{2017}-1}{-\dfrac{1}{7}-1}=\dfrac{\left(-\dfrac{1}{7}\right)^{2017}}{-\dfrac{8}{7}}\)
8 tháng 1 2021
Ta có: \(S=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2014}\)
\(\Leftrightarrow\dfrac{-1}{7}\cdot S=\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+\left(-\dfrac{1}{7}\right)^3+...+\left(-\dfrac{1}{7}\right)^{2015}\)
\(\Leftrightarrow S-\dfrac{-1}{7}\cdot S=\left(-\dfrac{1}{7}\right)^0-\left(-\dfrac{1}{7}\right)^{2015}\)
\(\Leftrightarrow\dfrac{8}{7}\cdot S=1+\dfrac{1}{7^{2015}}\)
\(\Leftrightarrow S=\left(1+\dfrac{1}{7^{2015}}\right):\dfrac{8}{7}=\dfrac{\left(1+\dfrac{1}{7^{2015}}\right)\cdot7}{8}\)
PX
0
HP
7 tháng 12 2015
7S=7+7^2+7^3+7^4+...+7^2016
=>7S-S=(7+7^2+7^3+7^4+...+7^2016)-(1+7+7^2+7^3+...+7^2015)
=>6S=7^2016-1
=>6S+1=7^2016-1+1=7^2016(đpcm)
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=\(\dfrac{(\dfrac{-1}{7})^{2017}-1}{\dfrac{-8}{7}}\)