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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 ~
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)..\left(1-\frac{1}{2016}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2015}{2016}\)
\(\Rightarrow A=\frac{1.2.3..2015}{2.3.4..2016}\)
\(\Rightarrow A=\frac{1}{2016}\)
\(\Rightarrow2A=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2014}\)
\(\Rightarrow2A-A=A=1-\left(\frac{1}{2}\right)^{2015}\)
Với B tương tự nhưng là lấy 3B
a) \(=\frac{3}{2}.\frac{4}{3}....\frac{100}{99}=\frac{100}{2}=50\)
a) =3/2 . 4/3 . 5/4 ...100/99
=\(\frac{3.4.5...100}{2.3.4..99}\)
=\(\frac{100}{2}\)
b) =
\(\left(1-\frac{1}{3}\right).\left(1-\frac{1}{5}\right).\left(1-\frac{1}{7}\right)...\left(1-\frac{1}{2}\right).\left(1-\frac{1}{4}\right).\left(1-\frac{1}{6}\right).\left(1-\frac{1}{120}\right)\)
\(=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{119}{120}=\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{118}{119}\right).\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{119}{120}\right)\)
\(=\frac{\left(2.4.6...118\right).\left(1.3.5...119\right)}{\left(3.5.7...119\right).\left(2.4.6...120\right)}=\frac{1}{120}\).
Đáp số: \(\frac{1}{120}\).
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2016}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2015}{2016}\)
\(=\frac{1.2.3....2015}{2.3.4....2016}\)
\(=\frac{1}{2016}\)