1.Tính tổng :
S = \(\left(\frac{-1}{7}\right)^o+\left(\frac{-1}{7}\right)^1+\left(\frac{-1}{7}\right)^2+...+\left(\frac{-1}{7}\right)^{2017}\)
Những câu hỏi liên quan
Tính tổng: \(S=\left(\frac{1}{7}\right)^0+\left(\frac{1}{7}\right)^1+\left(\frac{1}{7}\right)^2+...+\left(\frac{1}{7}\right)^{2014}\)
tính tổng \(S=\left(-\frac{1}{7}\right)^0+\left(-\frac{1}{7}\right)^1+\left(-\frac{1}{7}\right)^3+\left(-\frac{1}{7}\right)^4+.....+\left(-\frac{1}{7}\right)^{2007}\)
S=1-1/7-(1/7)^3-......-(1/7)^2017
49S=49-7-1/7-(1/7)^3-.,.....-(1/7)^2015
49S-S=48S=49-7-1-(1/7)^2017
48S=41-(1/7)^2017
S=41/48-(1/7)^2017/48
k nha
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Tính tổng: S= \(\left(-\frac{1}{7}\right)^0+\left(-\frac{1}{7}\right)^1+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2007}\)
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)
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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
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Nguyễn Huy Thắng giải giúp mjnk bài này vs
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Xem thêm câu trả lời
tính tổng S=\(\left(\frac{-1}{7}\right)^0+\left(\frac{-1}{7}\right)^1+\left(\frac{-1}{7}\right)^2+...+\left(\frac{-1}{7}\right)^{2007}\)
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)
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a) Tính \(S=\left(-\frac{1}{7}\right)^0+\left(-\frac{1}{7}\right)^1+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2017}\)
b) So \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\) với \(1\)
a) Tính tổng: \(S=\left(\frac{-1}{7}\right)^0+\left(\frac{-1}{7}\right)^1+\left(\frac{-1}{7}\right)^2+...+\left(\frac{-1}{7}\right)^{2007}\)
b) Chứng minh rằng : \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}
a)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)
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\(\left(\frac{1}{2}+\frac{2015}{2016}+\frac{2016}{2017}+1\right)\left(\frac{2105}{2016}+\frac{2016}{2017}+\frac{7}{22}\right)-\left(\frac{1}{2}+\frac{2015}{2016}+\frac{2016}{2017}\right)\left(\frac{2015}{2016}+\frac{2016}{2017}+\frac{7}{22}+1\right)\)
Tính S = \(\left(\frac{-1}{7}\right)^0+\left(-\frac{1}{7}\right)^1+\left(-\frac{1}{7}\right)^2+.....+\left(-\frac{1}{7}\right)^{2016}\)
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
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Tínha) left(1-frac{1}{2}right)left(1-frac{1}{3}right)...left(1-frac{1}{2018}right)b) left(1-frac{1}{3}right)left(1-frac{1}{6}right)left(1-frac{1}{10}right)left(1-frac{1}{15}right)...left(1-frac{1}{190}right)c) left(1+frac{7}{9}right)left(1+frac{7}{20}right)left(1+frac{7}{33}right)left(1+frac{7}{48}right)...left(1+frac{7}{2009}right)
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Tính
a) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2018}\right)\)
b) \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{190}\right)\)
c) \(\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)\left(1+\frac{7}{48}\right)...\left(1+\frac{7}{2009}\right)\)
a) =\(\frac{1}{2}.\frac{2}{3}.....\frac{2017}{2018}=\frac{1.2.....2017}{2.3.4.....2018}=\frac{1}{2018}\)
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