Chứng minh rằng:
a. \(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+...+\frac{99}{3^{100}}+\frac{100}{3^{101}}< \frac{1}{4}\)
b.\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
c.\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{1}{16}\)
d. \(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}< \frac{1}{36}\)
\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow3A+A=\left(...\right)+\left(...\right)\)
\(\Rightarrow4A=1-\frac{1}{3}+\frac{1}{3^2}-...-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3.4A=3-1+\frac{1}{3}-...-\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow12A+4A=\left(...\right)+\left(...\right)\)
\(\Rightarrow16A=3-\frac{101}{3^{99}}-\frac{100}{3^{100}}< 3\)
\(\Rightarrow A< \frac{3}{16}\)
Tính
B=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}}\)
a, Tính : \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b, Tính : \(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
c, Tính : \(\frac{\left(1+2+3+...+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
Tính giá trị của biểu thức \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}}\)
Tính giá trị của biểu thức \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}}\)
Tính \(A=\frac{1}{2.\sqrt{1}+1.\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+....+\frac{1}{100.\sqrt{99}+99.\sqrt{100}}\)
tính
\(\frac{\left(1+2+3+....+99+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63\times1,2-21\times3,6\right)}{1-2+3-4+....+99-100}\)
Tính M:
\(M=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+.....+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{100}}\)