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\(\frac{1}{100.101}+\frac{1}{101.102}+.......+\frac{1}{2010.2011}\)
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LM
5 tháng 8 2017
\(a,=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-0-0-0-...-0-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}\)
\(=\frac{4}{8}-\frac{1}{8}\)
\(=\frac{3}{8}\)
\(b,=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{49}+\frac{1}{49}-\frac{1}{16}\)
\(=1-0-0-0-...-0-\frac{1}{16}\)
\(=1-\frac{1}{16}\)
\(=\frac{15}{16}\)
\(c,\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\left(1-0-0-0-...-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\frac{50}{51}\)
\(=\frac{25}{17}\)
\(d,\)giống câu a tự làm nha mỏi tay quá.
BT
5 tháng 8 2017
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}.\)
=> \(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)
=> \(A=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
\(B=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{49.52}=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{49}-\frac{1}{52}\)
=> \(B=\frac{1}{4}-\frac{1}{52}=\frac{24}{104}=\frac{1}{26}\)
27 tháng 4 2017
A= 1/1-1/2+1/2-1/3+1/4-1/5+...+1/101-1/102
A=1-1/102=102/102-1/102=101/102
ý b thì chờ mình tí tìm cách lập luận đã nhé
DD
27 tháng 4 2017
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)
\(A=1-\frac{1}{102}\)
\(A=\frac{101}{102}\)
7 tháng 4 2015
a) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)(áp dụng quy tắc dấu ngoặt )
\(3A-A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^8}\)
\(3A-A=1+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{3^2}-\frac{1}{3^2}\right)+...+\left(\frac{1}{3^7}-\frac{1}{3^7}\right)-\frac{1}{3^8}\)
\(\Rightarrow2A=1+0+0...+0-\frac{1}{3^8}\)
\(2A=1-\frac{1}{3^8}\)
\(2A=\frac{3^8-1}{3^8}\)
\(A=\frac{3^8-1}{3^8}\div2=\frac{3^8-1}{3^8}.\frac{1}{2}=\frac{3^8-1}{3^8.2}\)
b) \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{100.101}\)
\(\Rightarrow B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\)(áp dụng quy tắc dấu ngoặt )
\(B=\frac{1}{1}-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-...-\left(\frac{1}{100}-\frac{1}{100}\right)-\frac{1}{101}\)
\(B=\frac{1}{1}-0-0-0...-0-\frac{1}{101}\)
\(B=\frac{1}{1}-\frac{1}{101}\)
\(B=\frac{100}{101}\)
PG
0
HH
0
TT
1
15 tháng 3 2016
\(D=\frac{\left(-1\right).\left(-1\right)}{1.2}.\frac{\left(-2\right).\left(-2\right)}{2.3}...\frac{\left(-101\right).\left(-101\right)}{101.102}\)
\(=\frac{\left(-1\right)\left(-1\right)\left(-2\right)\left(-2\right)...\left(-101\right)\left(-101\right)}{1.2.2.3...101.102}\)
\(=\frac{\left[\left(-1\right)\left(-2\right)...\left(-101\right)\right].\left[\left(-1\right).\left(-2\right)...\left(-101\right)\right]}{\left(1.2...101\right).\left(2.3...102\right)}\)
\(=\left(-1\right).\frac{-1}{102}\)
\(=\frac{1}{102}\)
Vì \(\frac{1}{102}>\frac{-1}{100}\)
Vậy\(D>\frac{-1}{100}\)
30 tháng 4 2016
Đặt \(A=\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+...+\frac{1}{100.101}\)
\(\Rightarrow A< \left(\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+...+\frac{1}{100.101}\right)+\left(\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+...+\frac{1}{99.100}\right)\)
\(\Rightarrow A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}+\frac{1}{100.101}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{101}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
Vậy \(\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+...+\frac{1}{100.101}< 2\) (đpcm)
Ta có:
\(\frac{1}{100.101}+\frac{1}{101.102}+...+\frac{1}{2010.2011}\)
\(=\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(=\frac{1}{100}-\frac{1}{2011}\)
\(=\frac{1911}{201100}\)
Ta có : \(\frac{1}{100.101}\)+ \(\frac{1}{101.102}\)+.....+\(\frac{1}{2010.2011}\)
= \(\frac{1}{100}\)- \(\frac{1}{101}\)+ \(\frac{1}{101}\)- \(\frac{1}{102}\)+.....+ \(\frac{1}{2010}\)-\(\frac{1}{2011}\)
= \(\frac{1}{100}\)- \(\frac{1}{2011}\) = .... Tự tính tiếp nhé bạn