1. Tính:
B=\(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+....+\dfrac{1000}{2^{1000}}\)
K
Khách
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16 tháng 9 2023
Tổng A có 1000 số hạng
A>(1001/1000^2+1000)*1000=1001*1000/1000*(1000+1)=1
A<(1001/1000^2)*1000=1001/1000=1+1/1000<1
Vậy 1<A<2 nên 1<A^2<4
NC
1
5 tháng 4 2017
\(\dfrac{1}{1}.\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{1}{3}+\dfrac{1}{3}.\dfrac{1}{4}+...+\dfrac{1}{999}.\dfrac{1}{1000}\\ =\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{999.1000}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}\\ =1-\dfrac{1}{1000}=\dfrac{999}{1000}\)
K
4
15 tháng 8 2017
\(D=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{1000}}\)
\(2D=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{1000}}\right)\)
\(2D=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{999}}\)
\(2D-D=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{999}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{1000}}\right)\)\(D=1-\dfrac{1}{2^{1000}}\)
NN
15 tháng 8 2017
\(D=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{1000}}.\)
\(2D=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{1000}}\right).\)
\(2D=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{999}}.\)
\(2D-D=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{999}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{1000}}\right).\)
\(D=1+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^2}\right)+...+\left(\dfrac{1}{2^{999}}-\dfrac{1}{2^{999}}\right)-\dfrac{1}{2^{1000}.}\)
\(D=1+0+0+...+0-\dfrac{1}{2^{1000}}.\)
\(D=1-\dfrac{1}{2^{1000}}.\)
Vậy.....
24 tháng 4 2021
Ta có: D\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2005}\right)\)
\(\Leftrightarrow D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2004}{2005}=\dfrac{1.2.3...2004}{2.3.4...2005}=\dfrac{1}{2005}\)
Ta có: \(E=\dfrac{1^2}{1.3}.\dfrac{2^2}{2.4}.\dfrac{3^2}{3.5}...\dfrac{999^2}{999.1000}.\dfrac{1000^2}{1000.1001}=\dfrac{\left(1.2.3.4...1000\right)\left(1.2.3.4...1000\right)}{\left(1.2.3....1000\right)\left(3.4.5....1001\right)}=\dfrac{2}{1001}\)
\(B=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{3^2}+.....+\dfrac{1000}{2^{1000}}\)
\(2B=2\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{3^3}+.....+\dfrac{1000}{2^{1000}}\right)\)
\(2B=1+1+\dfrac{3}{2^2}+......+\dfrac{1000}{2^{999}}\)
\(2B-B=\left(2+\dfrac{3}{2^2}+.....+\dfrac{1000}{2^{999}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+.....+\dfrac{1000}{2^{999}}\right)\)\(2B-B=2-\dfrac{1}{2}-\dfrac{2}{2^2}-\dfrac{1000}{2^{999}}\)
\(B=1-\dfrac{1000}{2^{999}}\)