K
Khách
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VH
1
26 tháng 6 2017
\(B=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{3^2}+.....+\dfrac{1000}{2^{1000}}\)
\(2B=2\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{3^3}+.....+\dfrac{1000}{2^{1000}}\right)\)
\(2B=1+1+\dfrac{3}{2^2}+......+\dfrac{1000}{2^{999}}\)
\(2B-B=\left(2+\dfrac{3}{2^2}+.....+\dfrac{1000}{2^{999}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+.....+\dfrac{1000}{2^{999}}\right)\)\(2B-B=2-\dfrac{1}{2}-\dfrac{2}{2^2}-\dfrac{1000}{2^{999}}\)
\(B=1-\dfrac{1000}{2^{999}}\)
20 tháng 8 2017
a) Vì A=\(\dfrac{15^{16}+1}{15^{17}+1}\) < 1
\(\Rightarrow\dfrac{15^{16}+1}{15^{17}+1}< \dfrac{15^{16}+1+14}{15^{17}+1+14}=\dfrac{15^{16}+15}{15^{17}+15}\) \(=\dfrac{15\left(15^{15}+1\right)}{15\left(15^{16}+1\right)}\) \(=\dfrac{15^{15}+1}{15^{16}+1}\)
Vậy A<B
20 tháng 8 2017
b) A=\(\dfrac{2006^{2007}+1}{2006^{2006}+1}>1\)
\(\Rightarrow\dfrac{2006^{2007}+1+2005}{2006^{2006}+1+2005}\)
= \(\dfrac{2006^{2007}+2006}{2006^{2006}+2006}\)
= \(\dfrac{2006\left(2006^{2006}+1\right)}{2006\left(2006^{2005}+1\right)}\)
= \(\dfrac{2006^{2006+1}}{2006^{2005}+1}\)
Vậy A>B
N
0
NH
23 tháng 4 2017
Ta có:
\(\dfrac{9}{n!}\)< \(\dfrac{n-1}{n!}\) = \(\dfrac{1}{(n-1)!} - \dfrac{1}{n!}\) với n > 10 (n thuộc Z)
\(\Rightarrow\) \(\dfrac{9}{10!} + \dfrac{9}{11!} + \dfrac{9}{12!} + ... +\dfrac{9}{1000!} \)
= \(\dfrac{1}{9!} - \dfrac{1}{10!} + \dfrac{9}{11!} + \dfrac{9}{12!} + ... +\dfrac{9}{1000!}\)
\(\Rightarrow\) \(\dfrac{1}{9!} - \dfrac{1}{10!} + \dfrac{1}{10!} - \dfrac{1}{11!} + \dfrac{1}{11!} - \dfrac{1}{12!} + ....\)
= \(\dfrac{1}{9!} - \dfrac{1}{1000!}\)
\(\Rightarrow \) \(\dfrac{9}{10!} + \dfrac{9}{11!} + ...+ \dfrac{9}{1000!} < \dfrac{1}{9!}\)
Chúc bn hc tốt.
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.....