Chứng minh rằng: \(\dfrac{9}{10!}+\dfrac{9}{11!}+\dfrac{9}{12!}+...+\dfrac{9}{1000!}< \dfrac{1}{9!}\)
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Những câu hỏi liên quan
23 tháng 4 2017
a) Đặt :
\(A=\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+.................+\dfrac{1}{100!}\)
Ta thấy :
\(\dfrac{1}{2!}=\dfrac{1}{1.2}\)
\(\dfrac{1}{3!}=\dfrac{1}{1.2.3}\)
\(\dfrac{1}{4!}=\dfrac{1}{1.2.3.4}< \dfrac{1}{3.4}\)
.....................................
\(\dfrac{1}{100!}=\dfrac{1}{1.2.3..........100}< \dfrac{1}{99.100}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...........+\dfrac{1}{99.100}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...........+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A< 1-\dfrac{1}{100}\)
\(A< \dfrac{99}{100}< 1\)
\(\Rightarrow A< 1\rightarrowđpcm\)
b) Đặt :
\(B=\dfrac{9}{10!}+\dfrac{9}{11!}+\dfrac{9}{12!}+.............+\dfrac{9}{1000!}\)
Ta thấy :
\(\dfrac{9}{10!}=\dfrac{10-1}{10!}=\dfrac{1}{9!}-\dfrac{1}{10!}\)
\(\dfrac{9}{11!}< \dfrac{11-1}{11!}=\dfrac{1}{10!}-\dfrac{1}{11!}\)
...................................................
\(\dfrac{9}{1000!}< \dfrac{1000-1}{1000!}=\dfrac{1}{999!}-\dfrac{1}{1000!}\)
\(\Rightarrow B< \dfrac{1}{9!}-\dfrac{1}{10!}+\dfrac{1}{10!}-\dfrac{1}{11!}+............+\dfrac{1}{999!}-\dfrac{1}{1000!}\)
\(B< \dfrac{1}{9!}-\dfrac{1}{1000!}\)
\(\Rightarrow B< \dfrac{1}{9!}\rightarrowđpcm\)
~ Chúc bn học tốt ~
23 tháng 6 2018
a, Ta có :
\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{1\cdot2\cdot3\cdot4}+...+\dfrac{1}{1\cdot2\cdot3\cdot...\cdot100}\\ < \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}< 1\\ \Rightarrow M< 1\\ \RightarrowĐpcm\)
VH
22 tháng 7 2018
N = \(\dfrac{1}{10^2}+\dfrac{1}{11^2}+\dfrac{1}{12^2}+...+\dfrac{1}{n^2}\)
= \(\dfrac{1}{10.10}+\dfrac{1}{11.11}+\dfrac{1}{12.12}+...+\dfrac{1}{n.n}\)
=> N < \(\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}+...+\dfrac{1}{\left(n-1\right).n}\)
=> N < \(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(=>N< \dfrac{1}{9}-\dfrac{1}{n}\)
=> N < \(\dfrac{1}{9}\)
Vậy N < \(\dfrac{1}{9}\)
YT
0
18 tháng 7 2017
\(B=\dfrac{9}{10!}+\dfrac{10}{11!}+...........+\dfrac{99}{100!}\)
Ta thấy :
\(\dfrac{9}{10!}=\dfrac{10-1}{10!}=\dfrac{1}{9!}-\dfrac{1}{10!}\)
\(\dfrac{10}{11!}< \dfrac{11-1}{11!}=\dfrac{1}{10!}-\dfrac{1}{11!}\)
..........................
\(\dfrac{99}{100!}< \dfrac{100-1}{100!}=\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(\Leftrightarrow B< \dfrac{1}{9!}-\dfrac{1}{10!}+\dfrac{1}{10!}-\dfrac{1}{11!}+...........+\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(\Leftrightarrow B< \dfrac{1}{9!}-\dfrac{1}{100!}\)
\(\Leftrightarrow B< \dfrac{1}{9!}\rightarrowđpcm\)
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.