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\(A=\dfrac{1}{2}+\dfrac{1}{2}^2+\dfrac{1}{2}^3+...+\dfrac{1}{2}\)10
\(= (\dfrac{1}{2}+\dfrac{1}{2}^9)+(\dfrac{1}{2}^2+\dfrac{1}{2}^8)+(\dfrac{1}{2}^3+\dfrac{1}{2}^7)+\dfrac{1}{2}\)10
\(= \dfrac{257}{512}+\dfrac{65}{256}+\dfrac{17}{128}+\dfrac{1}{1024}\)
\(=( \dfrac{514}{1024}+\dfrac{136}{1024})+(\dfrac{260}{1024}+\dfrac{1}{1024})\)
\(=\dfrac{650}{1024}+\dfrac{261}{1024}\)
\(=\dfrac{911}{1024}\)
Giải:
(1-1/22).(1-1/32).(1-1/42).....(1-1/102)
=3/2.2 . 8/3.3 . 15/4.4 . ... . 99/10.10
=1.3.2.4.3.5.....9.11/2.2.3.3.4.4.....10.10
=1.2.3.....9/2.3.4.....10 . 3.4.5....11/2.3.4.....10
=1/10.11/2
=11/20
Chúc bạn học tốt!
A= 1/2+1/2^2+1/2^3+....+1/2^10(1)
=> 2A = 1+1/2+1/2^2+...+1/2^9(2)
Lấy (2) - (1) ta có ;
=> A = 1-1/2^10
Vậy.................
A = 1/2 + 1/2^2 + 1/2^3 + ...+ 1/2^10
2A = 1 + 1/2 + 1/2^2 + ...+ 1/2^9
2A - A = 1 - 1/2^10
A = 1 - 1/2^10
Chúc học giỏi !!!
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)
\(A=1-\frac{1}{2^9}\)
A= 1/2+1/2^2+1/2^3+....+1/2^9(1)
=> 2A = 1+1/2+1/2^2+...+1/2^8(2)
Lấy (2) - (1) ta có ;
=> A = 1-1/2^19
Vậy.................
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2018}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2018.2019}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2018.2019}\right)\)
\(=2\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2019-2018}{2018.2019}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2019}\right)\)
\(=\frac{2017}{2019}\)
\(A=\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}}{\dfrac{2}{2}+\dfrac{2}{2^2}+...+\dfrac{2}{2^{10}}}=\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}}{2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)}=\dfrac{1}{2}\)
A = \(\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{10}}}{\dfrac{2}{2}+\dfrac{2}{2^2}+\dfrac{2}{2^3}+...+\dfrac{2}{2^{10}}}\)
= \(\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}}{2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)}\)
= \(\dfrac{1}{2}\)