1/1+2 + 1/1+2+3 + 1/1+2+3+4+.....+1/1+2+3+.....+1009
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LM
0
LM
1
S
14 tháng 4 2017
Ta có: \(\frac{1}{2^2}=\frac{1}{4}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)
................
\(\frac{1}{1009^2}< \frac{1}{1008.1009}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{1009^2}< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1008.1009}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1009^2}< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1008}-\frac{1}{1009}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1009^2}< \frac{1}{4}+\frac{1}{2}-\frac{1}{1009}=\frac{3}{4}-\frac{1}{1009}< \frac{3}{4}\)(đpcm)
SR
3
ML
20 tháng 7 2017
Sửa đề :
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x}+\frac{1}{x\left(x+1\right)}=\frac{504}{1009}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2018}\)
\(\Leftrightarrow x=2018-1\)
\(\Leftrightarrow x=2017\)
Vậy ...
DP
20 tháng 7 2017
Sửa đề \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{504}{1009}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2018}\)
\(\Leftrightarrow x=2017\)
VQ
0
27 tháng 2 2018
Ta có :
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{99^2}{99.100}\)
\(=\)\(\frac{1^2.2^2.3^2.....99^2}{1.2.2.3.3.4.....99.100}\)
\(=\)\(\frac{1^2.2^2.3^2.....99^2}{1^2.2^2.3^2.4^2.....99^2}.\frac{1}{100}\)
\(=\)\(\frac{1}{100}\)
CP
0
TH
0
A = \(\dfrac{1}{1+2}\) + \(\dfrac{1}{1+2+3}\)+ \(\dfrac{1}{1+2+3+4}\)+...+\(\dfrac{1}{1+2+3+...+1009}\)
Ta có công thức:
S = 1 + 2 + ...+ n = (n+1)\(\times\)n:2
Áp dụng công thức trên vào A ta có
A = \(\dfrac{1}{\left(2+1\right)\times2:2}\)+\(\dfrac{1}{\left(1+3\right)\times3:2}\)+...+\(\dfrac{1}{\left(1009+1\right)\times1019:2}\)
A = \(\dfrac{1}{2\times3:2}\)+\(\dfrac{1}{3\times4:2}\)+...+\(\dfrac{1}{1009\times1010:2}\)
A = \(\dfrac{2}{2\times3}\)+ \(\dfrac{2}{3\times4}\)+...+\(\dfrac{2}{1009\times1010}\)
A = 2 \(\times\)( \(\dfrac{1}{2\times3}\)+ \(\dfrac{1}{3\times4}\)+...+ \(\dfrac{1}{1009\times1010}\))
A = 2 \(\times\) ( \(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+...+ \(\dfrac{1}{1009}-\dfrac{1}{1010}\))
A = 2 \(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{1010}\))
A = 1 - \(\dfrac{2}{1010}\)
A = \(\dfrac{1008}{1010}\)
A = \(\dfrac{504}{505}\)