Giúp mk với
Câu 1:
Cho A = \(\dfrac{1}{\dfrac{99}{\dfrac{1}{2}+}}+\dfrac{2}{\dfrac{98}{\dfrac{1}{3}+}}+\dfrac{3}{\dfrac{97}{\dfrac{1}{4}+....}}+...+\dfrac{99}{\dfrac{1}{\dfrac{1}{100}}}\).
B =\(\dfrac{92}{\dfrac{1}{45}+}-\dfrac{1}{\dfrac{9}{\dfrac{1}{50}+}}-\dfrac{2}{\dfrac{10}{\dfrac{1}{55}+}}-\dfrac{3}{\dfrac{11}{\dfrac{1}{60}+....}}-...\dfrac{92}{\dfrac{100}{\dfrac{1}{500}}}\). Tính \(\dfrac{A}{B}\)
rút gọn
\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}}{\dfrac{1}{99}+\dfrac{2}{98}+...+\dfrac{99}{1}}\)
Cmr : \(\dfrac{1}{3}\) - \(\dfrac{2}{3^2}\) +\(\dfrac{3}{3^3}\) - \(\dfrac{4}{3^4}\) + ...+\(\dfrac{99}{3^{99}}\) - \(\dfrac{100}{3^{100}}\)< \(\dfrac{3}{16}\)
a/ Rút gọn 2 biểu thức sau: \(E=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}\)và \(F=\dfrac{94-\dfrac{1}{7}-\dfrac{2}{8}-\dfrac{3}{9}-...-\dfrac{94}{100}}{\dfrac{1}{35}+\dfrac{1}{40}+\dfrac{1}{45}+...+\dfrac{1}{500}}\)
b/ Tính E - 2F
CMR \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
1/S=\(\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{4}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)\)
2/B=\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2007}\right)\)
3/C=\(\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\)
1 CM
a, \(\left(\dfrac{1}{1}+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{2n-1}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2n}\right)=\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{2n}\)( n∈Z)
b, \(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}=\dfrac{99}{50}-\dfrac{97}{49}+...+\dfrac{7}{4}-\dfrac{5}{3}+\dfrac{3}{2}\)
CMR\(\dfrac{1}{5}< \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}< \dfrac{2}{5}\)
Chứng minh rằng:
\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)