A = 1/2 + 1/4 + 1/8 + ... + 1/128
A = 1/2^1 + 1/2^2 + 1/2^3 + ... + 1/2^7
2A = 1 + 1/2 + 1/2^2 + ... + 1/2^6
2A - A = 1 - 1/2^7 = A
G = 5 - 5^2/1*6 5^2/6*11 - ... - 5^2/101*106
G = -5(-1 + 5/1*6 + 5/6*11 + ... + 5/101*106)
G = -5(-1 + 1 - 1/6 + 1/6 - 1/11 + ... + 1/101 - 1/106)
G = -1.(-1/106)
G = 1/106
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2A-A=1-\frac{1}{128}\)
\(A=\frac{127}{128}\)
D = 6/1*3 + 6/3*5 + 6/5*7 + ... + 6/2009*2011
D = 3(2/1*3 + 2/3*5 + 2/5*7 + ... + 2/2009*2011)
D = 3(1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2009 - 1/2011)
D = 3(1 - 1/2011)
D = 3*2010/2011 = 6030/2011
F = (1 + 1/2)(1 + 1/3)(1 + 1/4)...(1 + 1/100)
F = 3/2*4/3*5/4*...*101/100
F = 101/2
Vậy_
\(A=\frac{1}{2}+\frac{1}{4}+\cdot\cdot\cdot+\frac{1}{128}\)
\(\Rightarrow2A=1+\frac{1}{2}+\cdot\cdot\cdot+\frac{1}{64}\)
\(\Rightarrow A=1-\frac{1}{128}=\frac{127}{128}\)
\(\frac{2^2}{3}\cdot\frac{3^2}{2\cdot4}\cdot\cdot\cdot\cdot\cdot\frac{59^2}{58\cdot60}\)
\(=\frac{2\cdot2}{1\cdot3}\cdot\frac{3\cdot3}{2\cdot4}\cdot\cdot\cdot\cdot\cdot\frac{59\cdot59}{58\cdot60}\)
\(=\frac{\left(2\cdot3\cdot\cdot\cdot\cdot\cdot59\right)\cdot\left(2\cdot3\cdot\cdot\cdot\cdot\cdot59\right)}{\left(1\cdot2\cdot\cdot\cdot\cdot\cdot58\right)\cdot\left(3\cdot4\cdot\cdot\cdot\cdot\cdot60\right)}\)
\(=\frac{59\cdot2}{1\cdot60}=\frac{59}{30}\)
\(D=\frac{6}{1\cdot3}+\cdot\cdot\cdot+\frac{6}{2009\cdot2011}\)
\(=3\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdot\cdot\cdot+\frac{2}{2009\cdot2011}\right)\)
\(=3\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\cdot\cdot\cdot+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=3\cdot\left(1-\frac{1}{2011}\right)\)
\(=3\cdot\frac{2010}{2011}=\frac{6030}{2011}\)
\(\frac{1\cdot2\cdot3+2\cdot4\cdot6-3\cdot6\cdot9+5\cdot10\cdot15}{1\cdot3\cdot7+2\cdot6\cdot14-3\cdot9\cdot21+5\cdot15\cdot35}\)
\(=\frac{2\cdot\left(1\cdot3+4\cdot6-3\cdot3\cdot9+5\cdot5\cdot15\right)}{7\cdot\left(1\cdot3+4\cdot6-3\cdot3\cdot9+5\cdot5\cdot15\right)}\)
\(=\frac{2}{7}\)
\(\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\cdot\cdot\cdot\cdot\left(1+\frac{1}{100}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\cdot\cdot\cdot\cdot\frac{101}{100}\)
\(=\frac{3\cdot4\cdot\cdot\cdot\cdot\cdot101}{2\cdot3\cdot\cdot\cdot\cdot\cdot100}\)
\(=\frac{101}{2}\)
\(5-\frac{5^2}{1\cdot6}-\cdot\cdot\cdot-\frac{5^2}{101\cdot106}\)
\(=5\cdot\left(1-\frac{5}{1\cdot6}-\cdot\cdot\cdot-\frac{5}{101\cdot106}\right)\)
\(=-5\cdot\left(-1+1-\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{101}-\frac{1}{106}\right)\)
\(=-5\cdot\left(\frac{-1}{106}\right)\)
\(=\frac{5}{106}\)
