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HT
3
23 tháng 6 2017
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+..................+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+..................+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+.......+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+......+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{99}-\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+.......+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+......+\dfrac{1}{50}\right)\)
\(=\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+.......+\dfrac{1}{100}\)
LM
13 tháng 8 2016
A=\(\frac{1}{1.2}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{2017.2018}\)
A=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{2017}\)-\(\frac{1}{2018}\)
A=1-\(\frac{1}{2018}\)
A=\(\frac{2017}{2018}\)
13 tháng 8 2016
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2017.2018}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2017}+\frac{1}{2018}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2017}+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)
\(A=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2017}+\frac{1}{2018}\)
Đến đây bình thường ta nhóm 2 số vào với nhau nhưng ở đây có lẻ số hạng nên không nhóm được => đề sai
S
30 tháng 7 2018
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\right)\)
\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
\(=\frac{99}{100}-\frac{1}{2}\cdot\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)
15 tháng 12 2017
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1-\frac{1}{100}\)
\(B=\frac{99}{100}\)
PH
0
N
27 tháng 11 2018
\(\frac{1}{1.2}+\frac{1}{3.4}+....+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{100}\right)\)
\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-....-\frac{1}{50}=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
=> \(2013x.\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)=2013x.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)
=> \(2013x.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)=2012.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\Rightarrow2013x=2012\Rightarrow x=\frac{2012}{2013}\)
Vậy \(x=\frac{2012}{2013}\)
p/s: --trình bày sai sót mong bỏ qua
= 22.50.51.52 : 6 – 51.50 = 88400 – 2550 = 85850.
\(A=\left(2-1\right).2+\left(4-1\right).4+\left(6-1\right).6+...+\left(100-1\right).100\\ A=2^2-2+4^2-4+6^2-6+...+100^2-100\\ A=\left(2^2+4^2+...+100^2\right)-\left(2+4+...+100\right)\\ A=2^2\left(1+2^2+3^2+...+50^2\right)-\dfrac{\left(100+2\right).50}{2}\\ A=\dfrac{4.50.51.52}{6}-\dfrac{102.50}{2}=85850\)