Tính tống:
S= 1/2.4 + 1/4.6 + 1/6.8 +...+ 1/2018.2020
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
D
4
NH
30 tháng 7 2019
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2018}-\frac{1}{2020}\)
\(=\frac{1}{2}-\frac{1}{2020}=\frac{1009}{2020}\)
30 tháng 7 2019
\(A=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2018.2020}\)
\(\Leftrightarrow A=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2018.2020}\right)\)
\(\Leftrightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2018}-\frac{1}{2020}\right)\)
\(\Leftrightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2020}\right)=\frac{1}{2}.\frac{1009}{2020}\)
\(\Leftrightarrow A=\frac{1009}{4040}\)
Vậy : \(A=\frac{1009}{4040}\)
NT
1
29 tháng 4 2021
Sửa đề: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)
Ta có: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)
\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2018\cdot2020}+\dfrac{2}{2020\cdot2022}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2018}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)
\(=2\cdot\dfrac{505}{1011}\)
\(=\dfrac{1010}{1011}\)
1 tháng 5 2019
\(A=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2018.2020}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2020}\right)\)
\(A=\frac{1}{2}.\frac{1009}{2020}\)
\(A=\frac{1009}{4040}\)
1 tháng 5 2019
A=1/2.4+1/4.6+1/6.8+...+1/2018.2020
=1/2(1/2-1/4+1/4-1/6+...+1/2018-1/2020)
=1/2(1/2-1/2020)
=1/2.1009/2020
=1009/4040
LT
11 tháng 3 2020
\(B=\frac{3}{2.4}-\frac{5}{4.6}+\frac{7}{6.8}-\frac{9}{8.10}+...+\frac{2019}{2018.2020}\)
\(B=\frac{3}{2.1.2.2}-\frac{5}{2.2.2.3}+\frac{7}{2.3.2.4}-\frac{9}{2.4.2.5}+...+\frac{2019}{2.1009.2.1010}\)
\(B=\frac{1}{4.}.\left(\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+...+\frac{2019}{1009.1010}\right)\)
\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{3}{2}-\frac{5}{2}+\frac{5}{3}+\frac{7}{3}-\frac{7}{4}-\frac{9}{4}+\frac{9}{5}+...+\frac{2019}{1009}-\frac{2019}{1010}\right)\)
\(B=\frac{1}{4.}.\left(\frac{3}{1}-4+4-4+4-...+4-\frac{2019}{1010}\right)\)
\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{2019}{1010}\right)=\frac{1011}{4040}\)
3 tháng 5 2021
Gọi tổng cần tính là \(A\)
Ta có: \(A=\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{38.40}\)
\(\Rightarrow2A=\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{38.40}\)
\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{38}-\dfrac{1}{40}\)
\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{40}=\dfrac{19}{40}\)
\(\Rightarrow A=\dfrac{\dfrac{19}{40}}{2}=\dfrac{19}{80}\)
HB
1
HG
15 tháng 12 2016
A=1/2.4+1/4.6+........+1/100.102
A=1/2-1/4+1/4-1/6+.......+1/100-1/102
A=1/2-1/102
A=51/102-1/102
A=50/102
A=25/51
NX
0
14 tháng 6 2019
a,A =\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{199.200}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{199}-\frac{1}{200}\)
= 1-\(\frac{1}{200}\)
=\(\frac{199}{200}\)
14 tháng 6 2019
b, B=\(\frac{3}{2.4}+\frac{3}{4.6}+\frac{3}{6.8}+...+\frac{3}{2018.2020}\)
=3.(\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+..+\frac{1}{2018.2020}\))
=3(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+..+\frac{1}{2018}-\frac{1}{2020}\))
= 3.(\(\frac{1}{2}-\frac{1}{2020}\))
=\(\frac{6057}{2020}\)
\(S=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2018.2020}\)
\(S=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2018.2020}\right)\)
\(S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)
\(S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2020}\right)\)
Tự tính
S=1/2.4+1/4.6+1/6.8+...+1/2018.2020
S=1/2.(2/2.4+2/4.6+2/6.8+...+2/2018.2020)
S=1/2.(1-1/4+1/4-1/6+1/6-1/8+...+1/2018-1/2020)
S=1/2.(1-1/2020)
S=1/2.(2020/2020-1/2020)
S=1/2.2019/2020
S=2019/4040