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TQ
0
ND
0
24 tháng 10 2019
a) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n}\right)\\ =\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{n-1}{n}\\ =\frac{1}{n}\)
b) \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{n}\right)\\ =\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{n+1}{n}\\ =n+1\)
c) \(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\\ =\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot\frac{3\cdot5}{4^2}\cdot...\cdot\frac{\left(n-1\right)\left(n+1\right)}{n^2}\\ =\frac{\left[1\cdot2\cdot3\cdot...\cdot\left(n-1\right)\right]\cdot\left[3\cdot4\cdot5\cdot...\cdot\left(n+1\right)\right]}{\left(2\cdot3\cdot4\cdot...\cdot n\right)\left(2\cdot3\cdot4\cdot...\cdot n\right)}\\ =\frac{n+1}{2n}\)
d) \(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)...\left(1+\frac{1}{99\cdot101}\right)\\ =\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot...\cdot\frac{10000}{99\cdot101}\\ =\frac{2^2\cdot3^2\cdot...\cdot100^2}{1\cdot3\cdot2\cdot4\cdot...\cdot99\cdot101}\\ =\frac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot99\right)\left(3\cdot4\cdot...\cdot101\right)}\\ =\frac{2\cdot100}{101}\\ =\frac{200}{101}\)
S
27 tháng 7 2019
\(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+.....-\frac{1}{2^{99}}\Rightarrow2A+A=3A=\left(1-\frac{1}{2}+\frac{1}{2^2}-....-\frac{1}{2^{99}}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+......-\frac{1}{2^{100}}\right)=1-\frac{1}{2^{100}}=\frac{2^{100}-1}{2^{100}}\Rightarrow A=\frac{2^{100}-1}{3.2^{100}}\)
\(2,4B=2+\frac{1}{2}+\frac{1}{2^3}+.....+\frac{1}{2^{97}}\Rightarrow4B-B=3B=\left(2+\frac{1}{2}+....+\frac{1}{2^{97}}\right)-\left(\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)=2-\frac{1}{2^{99}}=\frac{2^{100}-1}{2^{99}}\Rightarrow B=\frac{2^{100}-1}{3.2^{99}}\)
\(3,C=\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-....-\frac{1}{2^{58}}\Rightarrow8C=4-\frac{1}{2}+\frac{1}{2^4}-.....-\frac{1}{2^{55}}\Rightarrow8C+C=9C=\left(4-\frac{1}{2}+\frac{1}{2^4}-....-\frac{1}{2^{55}}\right)+\left(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-....-\frac{1}{2^{58}}\right)=4-\frac{1}{2^{58}}=\frac{2^{60}-1}{2^{58}}\Rightarrow C=\frac{2^{60}-1}{9.2^{58}}\)
L
2
26 tháng 6 2018
B1 : S = 1 + 2 + 2^2 + 2^3 + ... + 2^2008 / 1 - 2^2009
Đặt A = 1 + 2 + 2^2 + 2^3 + ... + 2^2008
2A = 2 + 2^2 + 2^3 + 2^3 + 2^4 + ... + 2^2009
2A - A = ( 2 + 2^2 + 2^3 + 2^4 + ... + 2^2009 ) - ( 1 + 2 + 2^2 + 2^3 + ... + 2^2008 )
A = 2^2009 - 1
S = 2^2009 - 1 / 1 - 2^2009
S = -1