tinh nhanh: 1/6+1/24+1/60..+1/970200
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HT
1
KK
17 tháng 10 2017
1+1/3+1/6+1/12+1/48+1/80+1/160 = (1+1/3+1/6+1/12+1/24+1/48)+1/80+1/160
Tách bt thành 2 tổng:
Tổng thứ 1: A=1+1/3+1/6+1/12+1/24+1/48
2A = 2 + 2/3+1/3+1/6+1/12+1/24
A= (2+2/3+1/3+1/6+1/12+1/24)-(1+1/3+1/6+1/12+1/24+1/48)=1+2/3 - 1/48 = 79/48
1/80+1/160= 3/160
Ta có : 79/48 + 3/160 = 799/480
T
28 tháng 5 2018
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)
\(\Rightarrow3A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(\Rightarrow3A=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)
\(\Rightarrow3A=1+1-\frac{1}{32}\)
\(\Rightarrow3A=2-\frac{1}{32}\)
\(\Rightarrow A=\left(2-\frac{1}{32}\right):3=\frac{21}{32}\)
P/s : Đúng nha
TD
1
12 tháng 3 2020
\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{18\cdot19\cdot20}\)
\(A=\frac{1}{2}\cdot\frac{2}{1\cdot2\cdot3}+\frac{1}{2}\cdot\frac{2}{2\cdot3\cdot4}+\frac{1}{2}\cdot\frac{2}{3\cdot4\cdot5}+...+\frac{1}{2}\cdot\frac{2}{18\cdot19\cdot20}\)
\(A=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{18\cdot19\cdot20}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2.3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{2}-0-0-...-0-\frac{1}{380}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)\)
\(A=\frac{1}{2}\cdot\frac{189}{380}\)
\(A=\frac{189}{760}\)
2
0
DT
4 tháng 7 2017
e) 2+4+6+8....+24
số hạng
(24-2):2+1=10
tổng
(24+2)x10:2=130
đ\s....
TK
0
HV
1
HV
1
HN
9 tháng 1 2022
\( \begin{array}{l} \Leftrightarrow B=\frac{1}{2} .\left(\frac{1}{6} +\frac{1}{12} +\frac{1}{20} +...+\frac{1}{240}\right)\\ \Leftrightarrow B=\frac{1}{2} .\left(\frac{1}{2.3} +\frac{1}{3.4} +\frac{1}{4.5} +...+\frac{1}{15.16}\right)\\ \Leftrightarrow B=\frac{1}{2} .\left(\frac{1}{2} -\frac{1}{3} +\frac{1}{3} -\frac{1}{4} +...+\frac{1}{15} -\frac{1}{16}\right)\\ \Leftrightarrow \ B=\frac{1}{2} .\left(\frac{1}{2} -\frac{1}{16}\right)\\ \Leftrightarrow \ B=\frac{1}{2} .\frac{7}{16}\\ \Leftrightarrow \ B=\frac{7}{32} \end{array}\)
\(A=\dfrac{1}{6}+\dfrac{1}{24}+\dfrac{1}{60}+...+\dfrac{1}{970200}\)
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)
Ta có:
\(\dfrac{1}{1.2.3}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)\);
\(\dfrac{1}{2.3.4}=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)\);
...;
\(\dfrac{1}{98.99.100}=\dfrac{1}{2}\left(\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{99.100-2}{200.99}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{98998}{200.99}\right)=\dfrac{49499}{19800}\)