A=\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+....+\dfrac{1}{2015.2016}\) tìm A
K
Khách
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10 tháng 6 2023
Ta biến đổi \(A=\dfrac{2-1}{1.2}+\dfrac{4-3}{3.4}+...+\dfrac{2016-2015}{2016.2015}+\dfrac{2018-2017}{2017.2018}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{1}{2017}-\dfrac{1}{2018}\)
\(A=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2017}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)
\(A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2017}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)
\(A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2017}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1009}\right)\)
\(A=\dfrac{1}{1010}+\dfrac{1}{1011}+...+\dfrac{1}{2017}+\dfrac{1}{2018}\)
Lại có \(B=\dfrac{1}{1010.2018}+\dfrac{1}{1011.2017}+...+\dfrac{1}{2018.1010}\)
\(B=\dfrac{1}{3028}.\left(\dfrac{3028}{1010.2018}+\dfrac{3028}{1011.2017}+...+\dfrac{3028}{2018.1010}\right)\)
\(B=\dfrac{1}{3028}\left(\dfrac{1}{1010}+\dfrac{1}{2018}+\dfrac{1}{1011}+\dfrac{1}{2017}+...+\dfrac{1}{2018}+\dfrac{1}{1010}\right)\)
\(B=\dfrac{1}{3028}.2\left(\dfrac{1}{1010}+\dfrac{1}{1011}+...+\dfrac{1}{2018}\right)\)
\(B=\dfrac{1}{3028}.2A\) \(\Rightarrow\dfrac{A}{B}=1514\inℤ\). Ta có đpcm
B
2
1 tháng 8 2021
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\\ =\dfrac{1}{1}-\dfrac{1}{2020}=\dfrac{2019}{2020}\)
ML
2
26 tháng 4 2023
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{19\cdot20}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1-\dfrac{1}{20}=\dfrac{19}{20}\)
BD
26 tháng 4 2023
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+....+\dfrac{1}{19\cdot20}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{20}\)
\(A=1-\dfrac{1}{20}\)
\(A=\dfrac{19}{20}\)
16 tháng 8 2023
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{149.150}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{149}-\dfrac{1}{150}\)
\(A=\dfrac{1}{1}-\dfrac{1}{150}\)
\(A=\dfrac{150}{150}-\dfrac{1}{150}\)
\(A=\dfrac{149}{150}\)
11 tháng 9 2023
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{50}\)
\(A=\dfrac{49}{50}\)
9 tháng 5 2022
Áp dụng công thức \(\dfrac{1}{k\left(k+1\right)}=\dfrac{1}{k}-\dfrac{1}{k+1}\), ta có:
\(A=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{999}-\dfrac{1}{1000}\right)=1-\dfrac{1}{1000}=\dfrac{999}{1000}\)
AH
Akai Haruma
Giáo viên
15 tháng 4 2023
a.
$A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{1000-999}{999.1000}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}$
$=1-\frac{1}{1000}=\frac{999}{1000}$
AH
Akai Haruma
Giáo viên
15 tháng 4 2023
b.
$5B=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+....+\frac{5}{495.500}$
$=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+....+\frac{500-495}{495.500}$
$=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{495}-\frac{1}{500}$
$=1-\frac{1}{500}=\frac{499}{500}$
$\Rightarrow B=\frac{499}{500}: 5= \frac{499}{2500}$