Tính : \(\left(-1\right)+\left(-1\right)^2+\left(-1\right)^3+...+\left(-1\right)^{2015}\)
Những câu hỏi liên quan
Tính
\(\frac{\left(-5\right).\left(-0,9\right)^{2^{ }}}{\left(1\frac{1}{2}\right)^{4^{ }}.\left(-3\frac{1}{3}\right)^{3^{ }}.\left(-1\right)^{2015}}\)
\(=\dfrac{\left(-5\right)\cdot0.81}{\left(\dfrac{5}{2}\right)^4\cdot\left(-\dfrac{10}{3}\right)^3\cdot\left(-1\right)}=\dfrac{-5\cdot0.81}{\dfrac{5^4\cdot10^3}{2^4\cdot3^3}}\)
\(=-4.05:\dfrac{5^7}{3^3\cdot2}=\dfrac{-81}{20}\cdot\dfrac{3^3\cdot2}{5^7}=\dfrac{-3^7\cdot2}{2^2\cdot5^8}=\dfrac{-3^7}{2\cdot5^8}\)
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\(\frac{2.1+1}{\left(1+1\right)^2}+\frac{2.2+1}{\left(2^2+2\right)^2}+\frac{2.3+1}{\left(3^3+3\right)^2}+....+\frac{2.2015+1}{\left(2015^2+2015\right)^2}+\frac{2.1016+1}{\left(2016^2+2016\right)^2}\)
tính tổng . ai giúp vs
Tính các tổng sau
\(a,S=1+\left(-2\right)+3+\left(-4\right)+...+\left(-2014\right)+2015\)
\(b,S=\left(-2\right)+4+\left(-6\right)+8+...+\left(-2014\right)+2016\)
\(c,S=1+\left(-3\right)+5+\left(-7\right)+...+2013+\left(-2015\right)\)
\(d,S=\left(-2015\right)+\left(-2014\right)+\left(-2013\right)+...+2015+2016\)
a) \(S=1+\left(-2\right)+3+\left(-4\right)+...+\left(-2014\right)+2015\)
\(\Leftrightarrow S=\left(1-2\right)+\left(3-4\right)+....+\left(2013-2014\right)+2015\)
Vì từ 1 đến 2014 có 2014 số hạng => có 1007 cặp => Có 1007 cặp -1 và số 2015
\(\Rightarrow S=\left(-1\right)\cdot1007+2015\)
<=>S=-1007+2015
<=> S=1008
Tính A = \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2014}\right)\left(1-\frac{1}{2015}\right)\left(1-\frac{1}{2016}\right)\)
Ta có :
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2016}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2015}{2016}\)
\(A=\frac{2.3.4.....2015}{2.3.4.....2015}.\frac{1}{2016}\)
\(A=\frac{1}{2016}\)
Vậy \(A=\frac{1}{2016}\)
Chúc bạn học tốt ~
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\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)..\left(1-\frac{1}{2016}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2015}{2016}\)
\(\Rightarrow A=\frac{1.2.3..2015}{2.3.4..2016}\)
\(\Rightarrow A=\frac{1}{2016}\)
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k đúng cho mk nha
Xem thêm câu trả lời
Cho A=\(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2015}-1\right).\left(\dfrac{1}{2016}-1\right).\left(\dfrac{1}{2017}-1\right)\)
B=\(\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{2015}\right).\left(-1\dfrac{1}{2016}\right).\left(-1\dfrac{1}{2017}\right)\)
Tính M=A.B
\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2015}-1\right)\left(\dfrac{1}{2016}-1\right)\left(\dfrac{1}{2017}-1\right)\\ A=\left(-\dfrac{1}{2}\right).\left(-\dfrac{2}{3}\right).\left(-\dfrac{3}{4}\right)...\left(-\dfrac{2014}{2015}\right)\left(-\dfrac{2015}{2016}\right)\left(-\dfrac{2016}{2017}\right)\\ A=\dfrac{1.2.3.4...2014.2015.2016}{2.3.4...2015.2016.2017}=\dfrac{1}{2017}\)
\(B=\left(-1\dfrac{1}{2}\right)\left(-1\dfrac{1}{3}\right)\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{2015}\right)\left(-1\dfrac{1}{2016}\right)\left(-1\dfrac{1}{2017}\right)\\ B=\left(-\dfrac{3}{2}\right)\left(-\dfrac{4}{3}\right)\left(-\dfrac{5}{4}\right)...\left(-\dfrac{2016}{2015}\right)\left(-\dfrac{2017}{2016}\right)\left(-\dfrac{2018}{2017}\right)\\ B=\dfrac{3.4.5...2016.2017.2018}{2.3.4...2015.2016.2017}=\dfrac{2018}{2}=1009\)
\(M=A.B=\dfrac{1}{2017}.1009=\dfrac{1009}{2017}\)
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tinh \(G=\frac{\left(1+\frac{2015}{1}\right)+\left(1+\frac{2015}{2}\right)+...+\left(1+\frac{2015}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)+....+\left(1+\frac{1000}{2015}\right)}\)
tính
A=\(\left(\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2016}\right)\left(1+\frac{1}{2}+...+\frac{1}{2015}\right)\left(1+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)\)
tính tích:
\(\left(1-\frac{1}{2014}\right).\left(1-\frac{2}{2014}\right).\left(1-\frac{3}{2014}\right)...\left(1-\frac{2015}{2014}\right)\)
NHẤT ĐỊNH SẼ CÓ PHÂN SỐ \(1-\frac{2014}{2014}=0\)
NÊN tích dãy số đó là 0
tk nha
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Tính B=$\frac{1}{3} \frac{1}{6}\left(1 2\right) \frac{1}{9}\left(1 2 3\right) ... \frac{1}{6045}\left(1 2 3 ... 2015\right)$13 16 (1 2) 19 (1 2 3) ... 16045 (1 2 3 ... 2015)