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\(\dfrac{1}{1000}+\dfrac{1}{1002}+\dfrac{1}{1004}+...+\dfrac{1}{2000}< \dfrac{1}{2}\)
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Những câu hỏi liên quan
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
13 tháng 3 2023
A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\) + .....+ \(\dfrac{1}{1002^2}\)
A = \(\dfrac{1}{2^2.1^2}\) + \(\dfrac{1}{2^2.2^2}\) + \(\dfrac{1}{2^2.3^2}\)+......+\(\dfrac{1}{2^2.501^2}\)
A = \(\dfrac{1}{2^2}\) \(\times\)( \(1\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\)+.......+ \(\dfrac{1}{501^2}\))
ta có : \(\dfrac{1}{2^2}\) < \(\dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}\) < \(\dfrac{1}{2.3}\)
................
\(\dfrac{1}{501^2}\) < \(\dfrac{1}{500.501}\)
Cộng vế với vế ta được
\(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) +.....+ \(\dfrac{1}{501^2}\) < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.......+\dfrac{1}{500.501}\)
\(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) +.....+ \(\dfrac{1}{501^2}\) < \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}-\dfrac{1}{3}\)+.....+ \(\dfrac{1}{500}-\dfrac{1}{501}\)
\(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\)+......+ \(\dfrac{1}{501^2}\) < 1 - \(\dfrac{1}{501}\) < 1
=>A = \(\dfrac{1}{4}\) \(\times\) ( 1 + \(\dfrac{1}{2^2}\)+ \(\dfrac{1}{3^2}\)+.....+\(\dfrac{1}{501^2}\)) < \(\dfrac{1}{4}\) \(\times\)(1 + 1)
A < \(\dfrac{1}{4}\) \(\times\) 2
A < \(\dfrac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 1 2022
\(bx^2=ay^2\Rightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\)
\(\Rightarrow\left(\dfrac{x^2}{a}\right)^{1000}=\left(\dfrac{y^2}{b}\right)^{1000}=\left(\dfrac{1}{a+b}\right)^{1000}\)
\(\Rightarrow\dfrac{x^{2000}}{a^{1000}}=\dfrac{y^{2000}}{b^{1000}}=\dfrac{1}{\left(a+b\right)^{1000}}\)
\(\Rightarrow\dfrac{x^{2000}}{a^{1000}}+\dfrac{y^{2000}}{b^{1000}}=\dfrac{1}{\left(a+b\right)^{1000}}+\dfrac{1}{\left(a+b\right)^{1000}}=\dfrac{2}{\left(a+b\right)^{1000}}\)
N
0
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
23 tháng 6 2018
a, Ta có :
\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{1\cdot2\cdot3\cdot4}+...+\dfrac{1}{1\cdot2\cdot3\cdot...\cdot100}\\ < \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}< 1\\ \Rightarrow M< 1\\ \RightarrowĐpcm\)
T
6 tháng 4 2017
đặt \(P=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2001}-\dfrac{1}{2002}\\ Q=\dfrac{1}{1002}+...+\dfrac{1}{2002}\)
ta có:
\(P=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2001}-\dfrac{1}{2002}\\ \Rightarrow P=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2001}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2002}\right)\\ \Rightarrow P=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2002}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2002}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2002}\right)\)\(\Rightarrow P=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2002}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2002}\right)\\ \Rightarrow P=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2002}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1001}\right)\\ \Rightarrow P=\dfrac{1}{1002}+...+\dfrac{1}{2002}\\ \Rightarrow P=Q\)\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2001}-\dfrac{1}{2002}=\dfrac{1}{1002}+...+\dfrac{1}{2002}\left(đpcm\right)\)
1 tháng 4 2018
\(\dfrac{x-1016}{1001}+\dfrac{x-13}{1002}+\dfrac{x+992}{1003}=\dfrac{x+995}{1004}+\dfrac{x-7}{1005}+1\)
<=>\(\dfrac{x-1016}{1001}-1+\dfrac{x-13}{1002}-2+\dfrac{x+992}{1003}-3=\dfrac{x+995}{1004}-3+\dfrac{x-7}{1005}-2\)
<=>\(\dfrac{x-2017}{1001}+\dfrac{x-2017}{1002}+\dfrac{x-2017}{1003}=\dfrac{x-2017}{1004}+\dfrac{x-2017}{1005}\)
<=>\(\left(x-2017\right)\left(\dfrac{1}{1001}+\dfrac{1}{1002}+\dfrac{1}{1003}-\dfrac{1}{1004}-\dfrac{1}{1005}\right)=0\)
vì 1/1001+1/1002+1/1003-1/1004-1/1005 khác 0 nên x-2017=0<=>x=2017
vậy..........
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 :
\(\dfrac{9}{10!}=\dfrac{10-1}{10!}=\dfrac{1}{9!}-\dfrac{1}{10!}\)
\(\dfrac{9}{11!}< \dfrac{11-1}{11!}=\dfrac{1}{10!}-\dfrac{1}{11!}\)
...................................................
\(\dfrac{9}{1000!}< \dfrac{1000-1}{1000!}=\dfrac{1}{999!}-\dfrac{1}{1000!}\)
\(\Rightarrow B< \dfrac{1}{9!}-\dfrac{1}{10!}+\dfrac{1}{10!}-\dfrac{1}{11!}+............+\dfrac{1}{999!}-\dfrac{1}{1000!}\)
\(B< \dfrac{1}{9!}-\dfrac{1}{1000!}\)
\(\Rightarrow B< \dfrac{1}{9!}\rightarrowđpcm\)
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