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LD
26 tháng 3 2021
Ta có: \(\frac{1}{4}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{9}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
\(\frac{1}{16}=\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)
.............................................................
\(\frac{1}{10000}=\frac{1}{100\cdot100}< \frac{1}{99\cdot100}\)
\(\Rightarrow\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}< 1\)(đpcm)
Xin tk
26 tháng 3 2021
Ta có: +++.....+=++.......+
mà ++.......+ < ++.......+
++.......+ < 1-+-+.......+-
++.......+ < 1-
++.......+ < 1
30 tháng 10 2023
\(E=\dfrac{\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2002}-1\right)\left(\dfrac{1}{2003}-1\right)}{\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{9999}{10000}}\)
\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{100^2}\right)}\)
\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{100}{101}\cdot\dfrac{101}{102}\cdot...\cdot\dfrac{2002}{2003}}{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)}\)
\(=\dfrac{100}{2003}:\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\right)\)
\(=\dfrac{100}{2003}:\left(\dfrac{101}{2}\right)=\dfrac{100}{2003}\cdot\dfrac{2}{101}=\dfrac{200}{202303}\)
6 tháng 7 2016
A/5+10+15+...+1500
=5+10+15+...+1500 ta có:1500-5:5+1=300(số hạng)
=(5+1500)x300:2=225750
LT
1
21 tháng 6 2016
\(=\frac{1}{1.3}.\frac{1}{2.4}...\frac{1}{9.11}=\frac{1}{1.2.3^2...9^2.10.11}\)
21 tháng 7 2017
Ta có: \(VT=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
\(4VT=\dfrac{1}{2^2:2^2}+\dfrac{1}{4^2:2^2}+\dfrac{1}{6^2:2^2}+...+\dfrac{1}{100^2:2^2}\)
\(4VT=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)
Lại có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(...\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
\(\Rightarrow4VT-1< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)(*)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}\) (**)
Từ (*) và (**) \(\Rightarrow4VT< 2-\dfrac{1}{50}\)
\(\Rightarrow VT< \dfrac{1}{2}-\dfrac{1}{200}< VP\Rightarrow\) đpcm
b) Ta có: \(2VT=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\)
\(2VT+VT=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\)
\(3VT=1-\dfrac{1}{64}< 1\)
\(\Rightarrow VT< \dfrac{1}{3}\) (đpcm)
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
16 tháng 7 2019
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\)
\(A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)
\(A=\frac{\left(1.2.3.....99\right).\left(3.4.5.....101\right)}{\left(2.3.4.....100\right).\left(2.3.4.....100\right)}\)
\(A=\frac{1.101}{2.100}=\frac{101}{200}\)
16 tháng 7 2019
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}......\frac{9999}{10000}\)
\(A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)
\(A=\frac{1.2.3.4.....99}{2.3.4.5.....100}.\frac{3.4.5.6.....101}{2.3.4.5.....100}\)
\(A=\frac{1}{100}.\frac{101}{2}\)
\(A=\frac{101}{200}\)