Các câu hỏi tương tự
26 tháng 3 2016 lúc 16:34
Tính dãy số sau :
\(D=\frac{100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{98}{99}+\frac{99}{100}}\)
Tính Q=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{100}}{\frac{100-1}{1}+\frac{102-2}{2}+...+\frac{100-99}{99}}\)
Tính nhanh :
A = \(\left(\frac{2}{3}+\frac{3}{4}+....+\frac{99}{100}\right)\cdot\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+....+\frac{98}{99}\right)-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\cdot\left(\frac{2}{3}+\frac{3}{4}+...+\frac{98}{99}\right)\)
Cm:\(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}=2\)
So sánh:
a) \(\frac{-3}{1.3}+\frac{-3}{3.5}+...+\frac{-3}{97.99}\)và \(\frac{49}{-20}\)
b)\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}và\frac{99}{202}\)
c)\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}và\frac{99}{100}\)
Chứng minh rằng
\(100\)\(-(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100})=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)\(\frac{99}{100}\)
A=\(\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+....+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{99}+\frac{1}{100}}\)
B=\(\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-....-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+....+\frac{1}{500}}\)
\(I=\frac{2^2}{2^2-1}.\frac{3^2}{3^2-1}.\frac{4^2}{4^2-1}........\frac{99^2}{99^2-1}.\frac{100^2}{100^2-1}\)
tính
\(E=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)